Question

a bowling ball is dropped from a height of 24 feet. write a function that gives the height h (in feet) of the bowling ball after t seconds

Answer 1

Answer:

$h=16t^2$

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32 ft/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow h=ut+\frac{1}{2}32\times t^2\\\Rightarrow h=ut+16t^2$

Here, u = 0 so,

$h=0t+16t^2\\\Rightarrow h=16t^2$

The equation is $h=16t^2$

$24=16t^2\\\Rightarrow t=\sqrt{\frac{24}{16}}\\\Rightarrow t=1.22\ s$

Time taken by the ball to hit the ground is 1.22 seconds

Answer 2

Gravity pulls at 32 feet per second so it would take .75sec to hit the ground now make an equation that can make this applicable so after .75 second T it would be at 0 feet H