Question

A cannon ball is shot straight upward with a velocity of 72.50 m/s. How high is the cannon ball above the ground 3.30 seconds after it is fired? (Neglect air resistance.)

Answer 1

Answer:

Explanation:

Given

Cannon is fired with a velocity of $u=72.50\ m/s$

Using Equation of motion

$y=ut+\frac{1}{2}at^2$

where

$y=displacement$

$u=initial\ velocity$

$a=acceleration$

$t=time$

after time $t=3.3 s$

$y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2$

$y=239.25-53.36$

$y=185.89\ m$

So after 3.3 s cannon ball is at a height of 185.89 m