The boy on the tower of height h = 20 m in the figure below throws a ball a distance of x = 52 m, as shown. At what speed, in m/s, is the ball thrown?
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To determine the speed at which the ball is thrown, we need to calculate the initial velocity of the ball.
To find the initial velocity, we can use the equation of motion in the vertical direction:
v^2 = u^2 + 2as
where:
v = final velocity (which is 0 m/s at the topmost point)
u = initial velocity
a = acceleration due to gravity (-9.8 m/s^2, assuming downward as negative)
s = displacement (in this case, the height of the tower, h)
Since the final velocity is 0, the equation becomes:
0^2 = u^2 + 2(-9.8)(20)
Simplifying the equation, we have:
0 = u^2 - 392
Rearranging the equation, we can find the value of u:
u^2 = 392
Taking the square root of both sides:
u = √392 ≈ 19.80 m/s
Therefore, the ball is thrown with an initial velocity of approximately 19.80 m/s.