Question

A cannon is fired with an initial horizontal velocity of 20 m/s and initial vertical velocity of 25 m/s. After 3s in the air, the canon hits its target. How far away was the canon from its Target?

Answer 1

140.16m

Step-by-step explanation:

To get the distance the cannon is away from its target, we will apply the equation of motion;

S = ut + 1/2gt²

u is the resultant initial velocity

g is the acceleration due to gravity

t is the time taken.

Given

g = 9.8m/s²

t = 3s

u = √20²+25²

u = √400+625

u = √1025

u = 32.02m/s

Substitute the values into the formula and get the distance S;

S = 32.02(3)+ 1/2(9.8)3²

S = 96.06+44.1

S = 140.16m

Hence the cannon was 140.16m away from its target