The boy on the tower of height h = 20 m in the figure below throws a ball a horizontal distance of x = 56 m. At what speed, in m/s, is the ball thrown?
To find the speed at which the ball is thrown, we need to use the principles of projectile motion. We can break down the motion into horizontal and vertical components.
In this scenario, the ball is thrown horizontally, so its initial vertical velocity is zero (since there is no vertical motion initially). The only force acting on the ball is gravity, which only affects the vertical motion.
We can use the equation for vertical motion to find the time it takes for the ball to reach the ground:
h = (1/2) * g * t^2
Here, h is the height of the tower (20 m) and g is the acceleration due to gravity (9.8 m/s^2). Since we want to find the time it takes for the ball to reach the ground, we set h equal to zero:
0 = (1/2) * 9.8 * t^2
Simplifying the equation:
0 = 4.9 * t^2
This equation has two solutions: t = 0 and t = 2. The time cannot be zero as the ball needs time to reach the ground. Therefore, t = 2 s.
Next, we can find the horizontal velocity (Vx) of the ball using the formula:
Vx = x / t
Given that x (horizontal distance) is 56 m and t (time) is 2 s, we substitute these values into the equation:
Vx = 56 / 2
Vx = 28 m/s
Therefore, the speed at which the ball is thrown is 28 m/s.