Answered

A ball with mass 0.13 kg is thrown upward with initial velocity 20 m/s from the roof of a building 30 m high. neglect air resistance. (a computer algebra system is recommended. use g = 9.8 m/s2 for the acceleration due to gravity. round your answers to one decimal place.) (a) find the maximum height above the ground that the ball reaches.

Answer :

calculista
We need to apply the following equation

Vy ^2 = Voy^2 + 2 ay (y - yo)

Where Vy is the final speed of the ball (which will be equal to zero at maximun height)

Voy is the initial speed 

a is the acceleration (equal to -9.8 m/s2 ,as it goes against the upward movement)

yo is the initial reference position ( yo = 30m from the ground)

y = yo + [(-Voy^2)/ 2ay]  = 30 m + -(20[m/s])^2 / -2(9.8[m/s^2])

y = 30[m] + 20.408 [m] = 50.408 m = 50.4 m

Other Questions