A ball is launched directly upward from ground level with aninitial speed of 19

Question

A ball is launched directly upward from ground level with aninitial speed of 19 m/s. (Air resistanceis negligible.) (a) How long is the ball in the air?
s

(b) What is the greatest height reached by the ball?
m

(c) How many seconds after launch is the ball 4 m above the release point?
s after being thrown (first time)
s after being thrown (second time) (a) How long is the ball in the air?
s

(b) What is the greatest height reached by the ball?
m

(c) How many seconds after launch is the ball 4 m above the release point?
s after being thrown (first time)
s after being thrown (second time)

Explanation / Answer

Given that the initial speed of the ball is u =19 m/s ------------------------------------------------------- At the maximum height H the final velocity of the ball is v =0 m/s Let time taken to reach the maximum height is t From the equation of kinematics                  v = u - gt                  t = u /g                    = 19 m/s / 9.8m/s2                    = 1.93 s Then time taken by the ball in air is T = 2t =3.9 s (b) From the equation of kinematics                    S= ut + (1/2)at2                    H = ut - (1/2)gt2                      = ------------ m (c) Let the height h = 4 m      The velocity of the ball at height his                  v2 = u2 + 2ah solve for V = ------- m      Time taken to reach the height his                     t1 = (V - u ) / -g                       = ------- sec The time taken to reach height h                   t2 = 2V / g                        =--------sec Total time taken by the ball reach the height second time is T= t1 + t2                                                                                             =--------------- sec                          = ------------ m (c) Let the height h = 4 m      The velocity of the ball at height his                  v2 = u2 + 2ah solve for V = ------- m      Time taken to reach the height his                     t1 = (V - u ) / -g                       = ------- sec The time taken to reach height h                   t2 = 2V / g                        =--------sec Total time taken by the ball reach the height second time is T= t1 + t2                                                                                             =--------------- sec    

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