Question

A ball is tossed vertically upward from a window, height 50.0 m above the ground. The initial speed of the ball is 25.0 m/s. The ball goes up and comes down, landing on the ground at the base of the building. Determine the following:

a) The time it took the ball to reach its maximum height above ground.

b) The maximum height above the ground.

c) The time the ball returns to its initial height.

d) The velocity of the ball at the time for part c)

e) The velocity and height of the ball above ground at time t = 6.00 s.

f) The total time that the ball is in the air.

g) The impact velocity of the ball with the ground.

Answer 1

along vertical

initial velocityvoy = 25 m/s

acceleration ay = -g = -9.8 m/s^2

initial position yo = 50 m

a)

at maximum height velocity vy = 0

vy = voy + ay*t

0 = 25 - 9.8*t

time = 2.55 s <<<<<------ANSWER

(b)

final position y = ymax

2*ay*(y - yo )= vy^2 - voy^2

-2*9.8*(ymax - 50) = 0^2 - 25^2

maimum height = 82 m

==========================

c)

after returning to initial position y = 50

y - yo = voy*t + (1/2)*ay*t^2

50 - 50 = 25*t - (1/2)*9.8*t^2

time t = 5.1s

===================================

d)

vy = voy + ay*t

vy = 25 - (9.8*5.1)

vy = -25 m/s

=========================

e)

vy = voy + ay*t

vy = 25 - 9.8*6

vy = -33.8 m/s

y - yo = voy*t + (1/2)*ay*t^2

y - 50 = 25*6 - (1/2)*9.8*6^2

height from ground y = 23.6 m

========================

f)

after returning to ground y = 0

y - y0 = voy*t + (1/2)*ay*t^2

0 - 50 = 25*t - (1/2)*9.8*t^2

4.9*t^2 - 25t - 50 = 0

a = 4.9     b = -25      c = -50

t = -b +sqrt(b^2 - 4ac)/(2a) and t = -b -sqrt(b^2 - 4ac)/(2a)

t = (25 + sqrt(25^2+(4*4.9*50))) /(2*4.9) = 6.64 s    and t = (25 - sqrt(25^2+(4*4.9*50))) /(2*4.9) = -1.54 s

t = 6.64 s

======================

g)

vy = voy + ay*t

vy = 25 - 9.8*6.64

vy = -40.1 m/s

=========================

DONE please check the answer. any doubts feel free to ask