A cannonball is shot (from ground level) with an initial horizontal velocity of

ID: 1593062 • Letter: A

Question

A cannonball is shot (from ground level) with an initial horizontal velocity of 36 m/s and an initial vertical velocity of 21 m/s.

What is the initial speed of the cannonball?

m/s

What is the initial angle of the cannonball with respect to the ground?

What is the maximum height the cannonball goes above the ground?

How far from where it was shot will the cannonball land?

What is the speed of the cannonball 2.8 seconds after it was shot?

m/s

How high above the ground is the cannonball 2.8 seconds after it is shot?

1) ux = 36 m/s

uy = 21 m/s

speed = sqrt(ux^2 + uy^2)

= sqrt(36^2 + 21^2) = 41.68 m/s

2) @ = tan^-1(uy / ux) = 30.26 deg

3) maximum height is when vertical velocity will become zero.

using v^2 - u^2 = 2ad

0^2 - 21^2 = 2(-9.81)H

H = 22.48 m

4) for time period using v = u + at

-21 = 21 - 9.81t

t = 4.28 s

distance = 36 x 4.28 = 154.13 m

5. after 2.8 s

vx =ux = 36 m/s

vy = 21 - 9.81(2.8) = - 6.5 m/s

speed = sqrt(36^2 + 6.5^2) = 36.58 m/s

6) h = uy*t + ay*t^2 /2

h = 21*2.8 + (-9.81 * 2.8^2 /2 )

h = 20.34 m

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.