Question

A shell is launched with a velocity of 100 m/s at an angle of 30.0° above horizontal from a point on a cliff 50.0 m above a level plain below. How far from the base of the cliff does the shell strike the ground? There is no appreciable air resistance, and g = 9.80 m/s2 at the location of the cliff.

Answer 1

Answer:

The shell strikes the ground 963 m far from the base of the cliff.

Explanation:

Projectile Motion

Suppose an object is launched from a height yo from the ground level at an initial speed vo and an angle $\theta$ respect to the horizontal reference, the height of the object at any time t is given by

$\displaystyle y=y_o+v_o\ sin\theta \ t-\frac{gt^2}{2}$

And the horizontal distance traveled by the object at any time t is

$\displaystyle X=v_o\ cos\theta\ t$

The shell is launched at 100 m/s and 30° from a point at a height of 50 m. We need to find the time at which the shell hits the ground after launch time. Replacing the values into the equation:

$\displaystyle 50+100\ sin\ 30^o\ t-\frac{9.8t^2}{2}=0$

Operating and rearranging

$\displaystyle -4.9\ t^2+50\ t+50=0$

This is a second-degree equation in t that has two real answers:

$\displaystyle t=11.125\ sec,\ t=-0.92\ sec$

Only the first answer will be used, since the time cannot be negative. Now we know the flight time, we find the horizontal distance from the base of the cliff where the shell hit the ground

$\displaystyle X=100\cdot cos\ 30^o\cdot 11.12$

$\displaystyle \boxed{X=963\ m}$

The shell strikes the ground 963 m far from the base of the cliff.