Pre-Calc Help?

Question

Stacey stands on the roof of a building and throws a ball upward. The ball travels according to the equation 
s(t) = -16t^2+64t + 60 
s(t) = height 
t = seconds 
1. How high does the ball travel? 
2. How many seconds does it take for the ball to hit the ground? 
3. How high above the ground is the ball when it is thrown?

Answer 1

(1)

Given ball travel distance s(t) = -16t² + 64t + 60

Maximum height travel by ball Hmax = ?

s(t) = -16t² + 64t + 60

Apply derivative with respect to t both sides.

ds/dt = -16*2t + 64

ds/dt = - 32t + 64

ds/dt = velocity = v

v = - 32t + 64

At maximum height  velocity v = 0

0 = - 32t_max+ 64

32t_max= 64

t_max = 64/32

t_max = 2 sec

at t_max = 2 sec ,  ball is at maximum height.

s(t) = H_max = 2 sec = -16t² + 64t + 60

Substitute t_max = 2 sec

H_max = -16(2)² + 64(2)+ 60

H_max = - 64 + 128 + 60

H_max = 124

Ball travels 124 ft height

Answer 2

(3)

Given ball travel distance s(t) = -16t² + 64t + 60

Th height above the ground when ball is thrown H = ?

H is also equal to roof height.

s(t) = -16t² + 64t + 60

When ball is thrown , t = 0

s  = H = -16(0)² + 64(0) + 60

H = 60 ft

Th height above the ground when ball is thrown is 60 ft.

Answer 3

(2)

Given ball travel distance s(t) = -16t² + 64t + 60

Total time to reach to ground t_total = ?

Time from maximum height to ground = t_gnd

Total time to reach to ground is sum of time from roof to maximum height and Time from maximum height to ground.

t_total = t_max + t_gnd

s(t) = -16t² + 64t + 60

Apply derivative with respect to t both sides.

ds/dt = -16*2t + 64

ds/dt = - 32t + 64

ds/dt = velocity = v

v = - 32t + 64

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 = - 32t_max+ 64

32t_max= 64

t_max = 64/32

t_max = 2 sec

at t_max = 2 sec ,  ball is at maximum height from ground.

s(t) = H_max = 2 sec = -16t² + 64t + 60

Substitute t_max = 2 sec

H_max = -16(2)² + 64(2)+ 60

H_max = - 64 + 128 + 60

H_max = 124

From equation of law of motion with initial conditions : s = ut_gnd + ½at²

When ball falls to ground , Ball starts at zero initial velocity , u =0

124 =  0 + ½(32)(t_gnd)²

(t_gnd)² = 124*2/32

(t_gnd)² = 7.75

t_gnd = 2.78 sec

t_total = t_max + t_gnd

t_total = 2 + 2.78

t_total = 4.78 sec.

Total time to reach to ground is 4.78 sec