Question

John drops a basketball from the top of a building. If the ball impacts the ground with a velocity of -19 m/s, how tall is the building?

Answer 1

Answer:

Assume that the acceleration downwards = gravitational field strength = 9.8 (2 s.f)

Assume that the mass of the ball is negligible and there is no air resistance.

Use the equation s=ut+1/2at2

s=height of building

t= time taken for the ball to drop

u=initial velocity ; this = 0, because when the ball is dropped it has no speed initially, however if the question had said projected instead of dropped then it must have an initial speed. In this scenario we assume it has 0 initial speed since the question says dropped.

Substitute values to equation :

s=(0*3)+(1/2*9.8*3^2)

s=(1/2*9.8*3^2)

s=44.1m

Answer 2

1.

we can use kinematics to find the final velocity

v_f = v_i + atv

f

=v

i

+at

v_f = 0 + 10*3 = 30 m/sv

f

=0+10∗3=30m/s

2.

Total time of the motion of ball T = 2 s

so we can use

T = \frac{2*v_i}{g}T=

g

2∗v

i

2 = \frac{2*v_i}{9.8}2=

9.8

2∗v

i

v_i = 9.8 m/sv

i

=9.8m/s

now maximum height will be

H = \frac{v_i^2}{2g}H=

2g

v

i

2

H = 4.9 mH=4.9m

also the final speed at which ball collide with the ground will be same as initial speed so its 9.8 m/s.

3.

velocity of object at any instant during free fall is given by

v_f = v_i + atv

f

=v

i

+at

After t = 3 s

v_f = 0 + 9.8*3 = 29.4 m/sv

f

=0+9.8∗3=29.4m/s

after t = 4 s

v_f = 0 + 9.8*4 = 39.2 m/sv

f

=0+9.8∗4=39.2m/s

height from which it is dropped is given by

d = \frac{v_f + v_i}{2}*td= 2v f

i∗td = \frac{0 + 39.2}{2}*4 = 78.4 md= 20+39.2 ∗4=78.4m

Hope this'll help you.