Question

A rock is thrown off a cliff at an angle of 290 with respect to the horizontal. The cliff is 58.5 m high. The initial speed of the rock is v0 = 39 m/s. Find the magnitude in m/s when the rock hits the ground. Ignore air resistance.

Enter your value with one(1) decimal places and no units. For example if your answer is 1.276, then enter 1.3.

Answer 1

We can use the kinematic equations to solve for the velocity of the rock when it hits the ground. We can split the velocity into horizontal and vertical components:

v0x = v0 * cos(290)

v0y = v0 * sin(290)

Next, we can use the equation for vertical motion with constant acceleration (g = 9.8 m/s^2 is the acceleration due to gravity) to find the final velocity:

vy = v0y - g * t

where t is the time it takes for the rock to fall 58.5 m. We can find t by using the equation for vertical motion with constant acceleration:

h = v0y * t - 0.5 * g * t^2

Solving for t:

t = sqrt(2 * h / g)

Now that we have t, we can find vy:

vy = v0y - g * t

Finally, we can find the magnitude of the velocity (vmag) by using the Pythagorean theorem:

vmag = sqrt(v0x^2 + vy^2)

vmag = sqrt(39^2 * cos(290)^2 + (39 * sin(290) - 9.8 * t)^2) = 55.8 m/s (rounded to one decimal place)

The magnitude of the velocity when the rock hits the ground is 55.8 m/s.