Question

A ball is thrown horizontally from the top of a 70.0-m building and lands 90.0 m from the base of the building. Ignore air resistance. What is the velocity just before it hits the ground?

Just give the magnitude in meters/second here.

Answer #1

We can solve this by applying equations of motion in horizontal and vertical directions separately

In vertical direction, initial velocity is zero and applying 2nd equation of motion i.e. s = ut + (1/2)at2

-70 = 0 + (1/2)*(-9.8)*t2

70 = 4.9*t2

t = 3.78s

Velocity in vertical direction just before hitting the ground, vy = at

**vy = -9.8*3.78 = 37.04 m/s**

Now, applying in horizontal direction, s = ut as there is no acceleration in horizontal direction

90 = vx*3.78

**vx = 23.81 m/s**

Magnitude of final velocity,

**v = 44.03 m/s**

Therefore, the velocity just before it hits the ground is
**44.03 m/s**

I hope i was able to answer your question. If you need anymore explanation, please ask in comments. I would be happy to help.

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