Question

If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 48ft/sec, its height after t seconds s(t)=64+48t-16t^2. What is the maximun height the ball reaches "in ft"? What is the velocity of the ball when it hits the ground in ft/sec. 

Answer 1

a)

Ball initial velocity u = 48 ft/sec

Building roof height x = 64 ft

Height s(t) = 64 + 48t - 16t²

ds/dt = 48 -16*2t

ds/dt = 48 - 32t

ds/dt = velocity = v

v = 48 - 32t

At maximum height  velocity v = 0

0 = 48 - 32t_max

32t_max= 48

t_max = 48/32

t_max = 1.5 sec

at t_max = 1.5 sec ,  ball is at maximum height.

s(t) = H_max = 1.5 sec = 64 + 48t - 16t²

Substitute t_max = 1.5 sec

H_max = 64 + 48(1.5) - 16(1.5)²

H_max = 64 + 72 - 36

H_max = 100

The maximum height the ball reaches is 100 ft.

Answer 2

(b)

Ball initial velocity u = 48 ft/sec

Building roof height x = 64 ft

Height s(t) = 64 + 48t - 16t²

Apply derivative with respect to t on both sides.

ds/dt = 48 -16*2t

ds/dt = 48 - 32t

ds/dt = velocity = v

v = 48 - 32t

Apply derivative with respect to t on both sides.

dv/dt = a = -32

a = - 32 ( when ball moves upwards )

a = + 32 ( when ball moves down wards )

At maximum height  velocity v = 0

0 = 48 - 32t_max

32t_max= 48

t_max = 48/32

t_max = 1.5 sec

at t_max = 1.5 sec ,  ball is at maximum height.

s(t) = H_max = 1.5 sec = 64 + 48t - 16t²

Substitute t_max = 1.5 sec

H_max = 64 + 48(1.5) - 16(1.5)²

H_max = 64 + 72 - 36

H_max = 100 ft

Ball includes maximum height and building height.So H_max = 100 ft

From equation of law of motion with initial conditions : s = s_o + ut - ½at²

Ball starts at zero initial velocity , u =0

When ball falls to ground , it has considerable final velocity v = ?

from equation of law of motion : v² - u² = 2aH_max

v² - 0² = 2(32)(100)

v² = 6400

v = √6400

v = 80 ft/s

When ball falls to ground , ball final velocity is 80 ft/s.