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.
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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