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.

Answer 1

Answer:

a) 256 ft

b) 32 ft/s

c) -32 ft/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32 ft/s²

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128^2}{2\times -32}\\\Rightarrow s=256\ ft$

Maximum height = 256 ft

When s = 240 ft

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -32\times 240+128^2}\\\Rightarrow v=32\ ft/s$

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

$v=u+at\\\Rightarrow 0=32-32t\\\Rightarrow t=\frac{-32}{-32}=1\ s$

Now, initial velocity will be considered as zero

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32\times 16+0^2}\\\Rightarrow v=32\ ft/s$

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