Question

If a ball is thrown vertically upward with a velocity of 128 ft/s, then its height after t seconds is s = 128t − 16t2.

(a) What is the maximum height reached by the ball?
(b) What is the velocity of the ball when it is 240 ft above the ground on its way up? (Consider up to be the positive direction.)
(c) What is the velocity of the ball when it is 240 ft above the ground on its way down?

Answer 1

a) 256 ft

b) 32 ft/s

c) -32 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

`v^2-u^2=2as\\\Rightarrow s=(v^2-u^2)/(2a)\\\Rightarrow s=(0^2-128^2)/(2* -32)\\\Rightarrow s=256\ ft`

Maximum height = 256 ft

When s = 240 ft

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* -32* 240+128^2)\\\Rightarrow v=32\ ft/s`

Speed of the ball at 240 ft is 32 ft/s while going up

Now this will be the velocity of the ball when it reaches 240 ft, which will be considered as the initial velocity

`v=u+at\\\Rightarrow 0=32-32t\\\Rightarrow t=(-32)/(-32)=1\ s`

Now, initial velocity will be considered as zero

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* 32* 16+0^2)\\\Rightarrow v=32\ ft/s`

Speed of the ball at 240 ft is -32 ft/s while going down