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.
