High School

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A student throws a softball horizontally from a dorm window 16.0 m above the ground. Another student, standing 12.0 m away from the dorm, catches the ball at a height of 1.55 m above the ground. What is the initial speed of the ball?

Answer :

The initial speed of the horizontally thrown ball is approximately 6.98 m/s.

The question pertains to calculating the initial speed of a horizontally thrown ball. Given that a student throws a softball horizontally from a dorm window 16.0 m above the ground and it is caught 12.0 m away and 1.55 m above the ground, we need to use the principles of projectile motion.

First, determine the vertical displacement: the ball drops from 16.0 m to 1.55 m, so the displacement (h) is 16.0 m - 1.55 m = 14.45 m. Using the equation of motion:

h = [tex]0.5 * g * t^2[/tex]

where g = 9.8 m/s^2 is the acceleration due to gravity, and t is the time of flight.

To find the time of flight (t):

14.45 = 0.5 * 9.8 * t^2

Simplifying, we get:

t^2 = 14.45 / 4.9 = 2.95

t = √(2.95) ~ 1.72 s

Next, we find the initial speed using the horizontal distance. The horizontal distance (d) is 12.0 m:

vx = d / t = 12.0 m / 1.72 s ~ 6.98 m/s

Therefore, the initial speed of the ball is approximately 6.98 m/s.

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