A ball is thrown upward with an initial velocity of 20 m/s. (a) how high above the thrower’s hand is the ball after 2 seconds? (b) how long is the ball in the air? a ball is thrown upward with an initial velocity of 20 m/s. (c) what is the ball’s instantaneous velocity after 1 second? (d) what is the ball’s displacement after 1 second? (e) what is the ball’s average velocity between t = 1 and t = 4 seconds?
The ball goes up vertically only with positive velocity. It reaches a maximum height where velocity is zero then the ball falls down with negative velocity.
Best place to start is by finding the time it takes to reach maximum height. v = u + at 0 = 20 - 9.8t 9.8t = 20 t = 20/9.8 t = 2 seconds So it takes 2 sec to reach max height then it will be 4 sec to return to it's starting point.
A. find the height at 2 sec. Δy = ut + (1/2)at² Δy = 20(2) + (1/2)(-9.8)(2)² Δy = 40 - 19.6 Δy = 20.4 m
B. we found that the ball is in the air for 4 seconds; 2 sec going up , 2 sec going down.
C. Instantaneous velocity at t = 1 sec ; will be going up +direction v = u + at v = 20 - 9.8(1) v = 10.2 m/s upward
D. displacement, Δy at t = 1 sec Δy = 20(1) - 4.9(1)² Δy = 20 - 4.9 Δy = 15.1 m
E. Average velocity = Δy / Δt We found that at t = 1 sec the ball is at 15.1 m and at t = 4 sec the ball has returned to starting point, 0 m. Δy = (0 - 15.1) = -15.1 m. Δt = (4 - 1) = 3 sec Average velocity = -15.1/3 = -5 m/s
