Question

A ball is thrown upward from the top of a building 160 ft high with an initial velocity of 64 ft/sec. Its height in feet above the ground is given by s(t)=−16t2+64t+160 where t

is in seconds.
a) When does the ball reach its maximum height?
b) What is its maximum height?
c) When does it hit the ground?
d) With what velocity does it hit the ground?
e) What was its acceleration at time t=1?

Answer 1

2 seconds

5.74165 seconds

-32 ft/s²

119.73 ft/s

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32 ft/s²

`s=-16t^2+64t+160`

`v=u+at\\\Rightarrow t=(v-u)/(a)\\\Rightarrow t=(0-64)/(-32)\\\Rightarrow t=2\ s`

The maximum height will be reached in 2 seconds

From the given equation

`s=-16(2)^2+64* 2+160\\\Rightarrow s=224\ ft`

The maximum height is 224 ft from the ground

`s=ut+(1)/(2)at^2\\\Rightarrow 224=0t+(1)/(2)* 32* t^2\\\Rightarrow t=\sqrt{(224* 2)/(32)}\\\Rightarrow t=3.74165\ s`

Time taken to fall from the maximum height is 3.74165 seconds

Time taken from the point in time the ball is thrown is 2+3.74165=5.74165 seconds

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* 32* 224+0^2)\\\Rightarrow v=119.73\ ft/s`

Velocity with which the ball will hit the ground is 119.73 ft/s

At t = 1

Acceleration of a body thrown is always i.e., a body in free fall is always 32 ft/s². Here as the ball is thrown up the it will be negative so -32 ft/s²