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 leve

Question

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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 leve

l 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.

Physics

1 answer:

Yuliya22 [2.4K] · Answer 1 · 2026-03-08T09:31:55+03:00

Answer:

The shell lands 963 m away from the cliff’s base.

Explanation:

Projectile Motion

Consider an object that is propelled from a height yo above ground level with an initial velocity vo and angle $\theta$ with respect to the horizontal. The height of this object at any time t can be calculated by

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

Meanwhile, the horizontal distance it covers at any instant t is given by

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

The shell is projected at 100 m/s at a 30° angle, starting from a height of 50 m. We must determine when the shell impacts the ground after being launched. Substituting the known values into the formula:

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

By simplifying and rearranging the equation

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

This leads to a quadratic equation in terms of t, which yields two real solutions:

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

Only the positive solution is considered since time cannot be negative. Now that we have the duration of the flight, we can compute how far from the cliff's base the shell hits the ground

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

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

The shell lands 963 m away from the cliff’s base.