Question

A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s.

What is the initial speed of the cannonball?

What is the initial angle θ 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 3.6 seconds after it was shot?

How high above the ground is the cannonball 3.6 seconds after it is shot?

Answer 1

Given,

vx = 33 m/s ; vy = 25 m/s

a)The intial velocity will be given by:

vi = sqrt (vx^2 + vy^2)

vi = sqrt (33^2 + 25^2) = 41.4 m/s

Hence, vi = 41.4 m/s

b)the angl wrt ground will be:

theta = tan^-1(vy/vx) = tan^-1(25/33) = 37.15 deg

Hence, theta = 37.15 deg

c)the cannonball can be modelled as a projectile, so the max height reached by it will be:

H = vy^2/2g

H = 25^2/2 x 9.8 = 31.89 m

Hence, H = 31.89 m

b)Let D be the distance where it lands.

D = vx t

we know from eqn of motion

v = u + at

at max height, v = 0

t = vy/g = 25/9.8 = 2.55 s

T = 2t = 2 x 2.55 = 5.1 s

D = vx t = 33 x 5.1 = 168.3 m

Hence, D = 168.3 m