Department of Natural Sciences Physics 1111 Quiz September 11, 006 Name SOLUTION A ball is thrown straight up and reaches its maximum height after.00 s. a. What is the acceleration of the ball after it leaves the hand and is rising? 9.81 m/s, downward b. What is the acceleration of the ball as it is falling? 9.81 m/s, downward c. What was the ball s initial velocity? V = 0 V = V 0 g t 0 = V 0 g t V 0 = g t = (9.81 m/s )(.00 s) = 19.6 m/s d. How high did the ball rise above the point of release? y = y 0 + V 0 t ½ gt y = 0 + (19.6 m/s)(.00 s) ½ (9.81 m/s ) (.00s) y = 19.6 m
Department of Natural Sciences Physics 1111 Quiz 3.1 September 15, 005 Name A stone is dropped from the edge of a cliff and it strikes the bottom of the cliff 3.50 s later. (a) How high is the cliff? y = 0.00 m y 0 =? V 0 = 0 m/s t = 3.50 s y = y 0 + V 0 t ½ gt 0 = y 0 ½ gt y 0 = ½ gt y 0 = 60.0 m (b) What is the stone s speed just before hitting? V = V 0 - gt V = (0 m/s) (9.81 m/s )(3.50 s) = -34.3 m/s V = 34.3 m/s (c) What is the stone s acceleration 1.00 s after it is dropped? Acceleration due to gravity, g = 9.81 m/s, downward
Department of Natural Sciences Physics 1111 Quiz 3 September 1, 007 Name SOLUTION A boy on a bridge throws a stone vertically downward with an initial speed of 14.7 m/s toward the river below. a. If the stone hits the water.00 s later, what is the height of the bridge above the water? V 0 = - 14.7 m/s t =.00 s y = 0 y = y 0 + V 0 t ½ g t 0 = y 0 + V 0 t ½ g t y 0 = - V 0 t + ½ g t y 0 = - (-14.7 m/s) (.00s) + ½ (9.81 m/s ) (.00s) y 0 = 49.0 m b. What is the velocity of the stone just before it hits the water? V = V 0 - g t V = (-14.7 m/s) - (9.81 m/s ) (.00s) = 34.3 m/s c. What is the acceleration of the stone at half distance between the bridge and the water? g = 9.81 m/s, downward
Department of Natural Sciences Clayton College & State University Physics 1111 Quiz January 6, 004 Name SOLUTION 1. Upon graduation, a joyful physics student throws his cap upward with an initial speed of 14.7 m/s. Given that its acceleration is 9.80 m/s (we neglect air resistance), a. How long does it take the cap to reach its highest point? V 0 = 14.7 m/s V = 0 m/s V = V 0 g t t = (V - V 0 )/ (- g) t = ( 0 m/s 14.7 m/s) / (- 9.80 m/s ) = 1.50 s y = y o + V o t ½g t b. What is the distance to its highest point? y = (0 m) + (14.7 m/s) (1.50 s) ½ (9.80 m/s ) (1.50 s) y o = 11.0 m y = y o + V o t ½g t y = 0 m y 0 = 0 m 0 = V o t ½g t 0 = t (V o ½g t ) c. What is the total time the cap is in the air, given that the cap returns to its original elevation?
0 = V o ½g t t = V 0 /g = (14.7 m/s) / (9.80 m/s ) = 3.00 s d. What is the acceleration of the cap at the highest point? Acceleration is constant over the whole motion: 9.80 m/s, downward. Department of Natural Sciences Department of Natural Sciences Physics 1111 Quiz January 5, 006 Name SOLUTION An airplane travels 80 m down the runway before taking off. If it starts from rest, moves with constant acceleration, and becomes airborne in 8.00 s, a. What is its speed, in m/s, when it takes off? x - x 0 = ½ (V + V 0 ) t x - x 0 = ½ V t V = ( x - x 0 )/ t V = (80 m) / (8.00 s) = 70.0 m/s b. What is its acceleration? V = V 0 + a t a = (V - V 0 )/ t a = (70.0 m/s) / (8.00 s) = 8.75 m/s c. What is the average velocity of the airplane during these 8 seconds? V av = (x - x 0 ) / t = (80 m) / (8.00 s) = 35.0 m/s
Department of Natural Sciences Physics 1111 Quiz 3 January 30, 008 Name SOLUTION A stone is thrown vertically upward with a speed of 18.0 m/s. a. What is the maximum height the stone reaches? V y = V y0 - g (y y 0 ) 0 = V y0 - g (y 0) V y0 = g y y = V y0 /( g) y = 16.5 m b. How long is required to reach this height? V y = V y0 - gt 0 = V y0 - gt V y0 = gt t = V y0 /g t = 1.83 s c. What is velocity and acceleration of the stone at the moment it reaches the heigest point? V y = 0 a = -g
V y = -18.0 m/s d. The stone is caught at the same elevation as it was thrown from. What is its velocity just before it is caught? Department of Natural Sciences Clayton College & State University Physics 1111 Quiz 3 June 13, 005 Name SOLUTION 1. A baseball is seen to pass upward by a window 5.0 m above the street with a vertical speed of 14.0 m/s. If the ball was thrown from the street, a. What was its initial speed? V = V 0 - g (y y 0 ) V 0 = V + g (y y 0 ) V 0 = (14.0 m/s) + (9.80 m/s )(5.0 m 0 m) V 0 = 6. m/s b. What altitude does it reach? V = V 0 - g (y y 0 ) 0 = V 0 - g (y y 0 ) (y y 0 ) = V 0 /( g) y = (6. m/s) /( x 9.80 m/s ) y = 35.0 m c. How much time is the ball in the air before it reaches the street level again?
y = y 0 + V 0 t - 1/ g t y = y 0 = 0 0 = V 0 t - 1/ g t = t ( V 0-1/ g t) V 0-1/ g t = 0 V 0 / g = t t = (6. m/s) / (9.80 m/s ) = 5.35 s Department of Natural Sciences Physics 1111 Quiz 3 June 11, 007 Name SOLUTION Coasting due east on your bicycle at 8.00 m/s, you encounter a sandy patch of road 7.0 m across. When you leave the sandy patch your speed has been reduced to 6.50 m/s. V 0 = 8.00 m/s V = 6.50 m/s x - x 0 = 7.0 m a. Assuming the bicycle slows with constant acceleration, what was its acceleration in the sandy patch? Give both magnitude and direction. V = V 0 + a (x - x 0 ) V - V 0 = a (x - x 0 ) (V - V 0 )/ ((x - x 0 )) = a a = ((6.50 m/s) (8.00 m/s) )/ ((7.0 m)) = -1.51 m/s (Negative value means that the
direction of acceleration is west) b. What time did it take you to cross the patch? V = V 0 + a t V - V 0 = a t t = (V - V 0 ) / a t = ((6.50 m/s) (8.00 m/s))/(-1.51 m/s ) = 0.993 s c. What was your average velocity on the sand? V av = (x - x 0 )/(t - t 0 ) = (7.0 m)/(0.993 s) = 7.5 m/s Department of Natural Sciences Clayton College & State University Physics 1111 Quiz 3 February 4, 009 Name SOLUTION A bullet is fired through a board 10.0 cm thick in such a way that the bullet s line of motion is perpendicular to the face of the board. If the initial speed of the bullet is 400 m/s and it emerges from the other side with a speed of 300 m/s, find a. Acceleration of the bullet as it passes through the board. V 0 = 400 m/s V = 300 m/s (x-x 0 ) = 10.0 cm = 0.100 m V = V 0 + a (x-x 0 ) a = (V - V 0 )/((x-x 0 ))
a = ((300 m/s) (400 m/s) )/((0.100 m)) = - 350,000 m/s = - 3.50 x 10 5 m/s b. The total time the bullet is in contact with the board. V = V 0 + at t =(V - V 0 )/a = ((300 m/s) (400 m/s))/( - 3.50 x 10 5 m/s ) =.86 x 10-4 s c. The average velocity of the bullet as it travels through the board. V av = x/ t = (0.100 m)/(.86 x 10-4 s) = 350 m/s Department of Natural Sciences Physics 1111 Quiz 3 February 3, 010 Name SOLUTION A geologist at the top of a deep crevasse cannot resist the temptation to hurl a gneiss rock down to the bottom. The rock has an initial upward speed of 10.0 m/s as it leaves the geologist s hand. V 10.0 m/ s 0 a. How high above the starting point will the rock rise? V 0 y 0 0 V V 0 0 V gy 0 g( y y0)
V0 gy y V 0 /(g) y = 5.10 m b. The rock hits the bottom 5.90 s after it is thrown. How deep is the crevasse? y 1 gt y 0 V 0 t y 0 (10.0 m/ s)(5.90 s) 1 (9.80 m/ s )(5.90 s) y = -11 m c. What is the speed of the rock at the instant just before the impact. V 0 V gt V (10.0 m/ s) (9.80 m/ s )(5.90 s) V = -47.8 m/s
c. Why are there two answers to (b)?