Question

Maybe it's because I'm over-analyzing it but I've wasted 20 sheets of paper trying to get the right answer for this problem, but I just don't get it

A shot put is released with a velocity of 12 m/s and stays in the air for 2.0s.


At what angle with the horizontal was it released?


What horizontal distance did it travel?

Answer 1

A shot put is released with a velocity of 12 m/s and stays in the air for 2.0s.

At what angle with the horizontal was it released?

What horizontal distance did it travel?"

For the vertical component,

v_initial = 12 sin x
v_final = 0
a = -9.8 m/s^2
t = 1s

Vf = Vi + at
0 = 12 sin x + -9.8(1)
-12 sin x = -9.8
12 sin x = 9.8
x = 54.75°

<< Angle to horizontal = 54.75° >>

Horizontal distance = V_x_component * t
d = 12 cos 54.75 * 2
<< d = 13.85 m >>

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