Question

A cannon is elevated at an angle of 30 degrees and fires a cannonball with a speed "v₁". The Cannon is on top of a cliff with a height of "h". Given v₁ and h, determine:

a) The speed of the cannoball when it is at its maximum height.

b) The speed of the cannonball just before hitting the ground.

c) The angle the cannonball's velocity makes with the horizontal just before hitting the ground.

d) The horizontal distance the ball travels before hitting the ground.

e) The time the cannonball is in the air.

Answer 1

Given,

= 30 Deg ; intial speed = v1 and height = h

the horizontal and vertical component of velocity will be:

v1x = v1 cos and v1y = v1 sin

a)Its vertical speed at max heigt will be zero. As the at maximum height the vertical speed of the projectile is zero, but horizontal remains in tact. So, v(hmax) = vix = v1 cos(30) = 0.866 v1

b)Let Vf be its velocity just before hitting the ground, From equation of motion we know v = u + at

In our case, v = Vf ; u = v1y and a = -g = 9.8

Vfy = v1y - gt

We need to know the time t as well. We know from equation of motion that,

Y = Yo + u t + 1/2 g t²

fir ground y = 0 y0 = h ;

4.9 t² - v1y t - h = 0

From the above quadratic eqn we can determine the time t. Putting that in expression of velocity we can get the magnitude of velocity.

(c)x comp of velocity did not change, So Vfx = v1x = v1 cos(30)

Vfy =  v1y - gt

= tan -1 (Vfy/Vfx)

(d) Horizontal distance will be given by

X = (final horizontal velocity) x time = v1 cos x t