Question

A cannon is shot at 48 m/s at an angle of 30 degrees to the horizontal off a cliff that is 78.4 meters tall. How fast will the cannonball be moving when it hits the ground at the bottom of the cliff?

Answer 1

Ans:-

Given data v= 48m/s,θ= 30deg

Horizontal                                                           vertical

Vix = v cos30 =41.57m/s                                                Viy = v sin30 =24m/s

ax = 0                                                                    ay = g = -9.81m/s^2

                                                                                Yf = -78.4m

We have to calculate final velocity

First find out time

t = ?

y = viy t + 1/ 2 at^2

-78.4 = (24)t + 1/ 2 (-9.81 )t^2

4.9t^2-24t = 78.4

t = 7.14s

The stone will reach the bottom of the cliff after 7.14seconds.

Now x = ?

x = vix*t

x = (41.57)(7.14)

x = 296.81m

The stone will hit 296.81 meters from the base of the cliff.

Now

vfx = vix = 41.57 m/s

vfy = ?

vfy = viy + at

vfy = 24 + (-9.81)(7.14)

vfy = -46.04m/ s

The horizontal component of velocity is +41.57 m/s and the vertical component of velocity is –46.04 m/s.so the final velocity vf = ?

Vf^2 = vfx^2 + vfy^2

Vf^2 = (41.57)^2 + (-46.04)^2

Vf^2 = 1728.07+2119.6816

Vf^2 = 3847.7516

Vf = 62.03m/ s

The final speed of the stone is 60 m/s.