Use the worked example above to help you solve this problem. A ball is thrown fr

Question

Use the worked example above to help you solve this problem. A ball is thrown from the top of a building with an initial velocity of 20.4 m/s straight upward, at an initial height of 50.4 m above the ground. The ball just misses the edge of the roof on its way down, as shown in the figure. Determine the time needed for the ball to reach its maximum height. Determine the maximum height. Your response differs from the correct answer by more than 100%. Determine the time needed for the ball to return to the height from which it was thrown, and the velocity of the ball at that instant. Determine the time needed for the ball to reach the ground Determine the velocity and position of the ball at t = 5.65 s.

Explanation / Answer

(A) at maximum height, v = 0

u = 20.4 m/s

a = - 9.8 m/s^2

v = u + a t

0 = 20.4 - 9.8t

t = 2.08 sec

(B) v^2 - u^2 = 2 a d

0^2 - 20.4^2 = 2(-9.8)(d)

d = 21.23 from the top of building

(C) t' = 2 t = 2 x 2.08 = 4.16 sec

v= 20.4 m/s downward or - 20.4 m/s

(d) y = - 50.4 m to reach the ground.

y = u t + a t^2 / 2

- 50.4 = 20.4t - 9.8 t^2 /2

4.9 t^2 - 20.4t - 50.4 = 0

t = 5.905 sec

(e) v = u + a t

v = 20.4 + ( - 9.81 x 5.65) = - 35.03 m/s .........Ans

dispalcement = (20.4 x 5.65) - (9.81 x 5.65^2 / 2) = - 41.3 m

so height from ground = 50.4 - 41.3 = 9.08 m .....Ans

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