A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale.

-What is the magnitude of the acceleration of the shell just before it strikes the ground?

Question

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale.

-What is the magnitude of the acceleration of the shell just before it strikes the ground?

Quizplus · Accepted Answer

The horizontal velocity is constant, so we can use the horizontal distance and time of flight to find the horizontal velocity.

d = vt
1330 m = v(t)
t = 1330 m/v

The vertical distance can be found using the equation:
y = vi(t) + (1/2)at^2
where vi = 0 and y = -80 m (taking upward as positive).

-80 m = (1/2)(a)(t^2)
a = - (2*80 m) / (t^2)

Substituting t from the first equation:
a = - (2*80 m) / ((1330 m/v)^2)

The negative sign indicates that the acceleration is downward (towards the ground).

Now, using the equation for range:
d = ((v^2)/g) * sin(2θ)

Since the shell was fired horizontally, θ = 0, so:
v = sqrt((d*g)/sin(2θ)) = sqrt((1330 m/s^2) / sin(0)) = 115.4 m/s

Therefore:
a = - (2*80 m) / ((1330 m/115.4 m/s)^2) = -9.8 m/s^2

The magnitude of this acceleration is 9.8 m/s^2, which corresponds to answer choice B.

A Shell Is Fired with a Horizontal Velocity in the Positive