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.)

Question

asked Feb 2, 2021 11.9k views

1 Answer

James World · Answer 1 · 2021-02-08T14:31:39+0000

Answer:

Step-by-step explanation:

Given

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

Using Equation of motion

y=ut+(1)/(2)at^2

where

y=displacement

$u=initial\ velocity$

a=acceleration

t=time

after time
t=3.3 s

y=72.50* 3.3-(1)/(2)* 9.8* (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