A ball is thrown horizontally at 10.0 m/s from the top of a hill50.0 m high. How far from the base of the cliff would the ball hit the ground?
45.0m
To find the horizontal distance from the base of the cliff where the ball hits the ground, we need to consider the horizontal velocity and the time of flight of the ball.
First, we need to find the time it takes for the ball to reach the ground. To do this, we can use the kinematic equation:
h = ut + (1/2)gt^2
Where:
h is the height of the cliff (50.0 m)
u is the initial vertical velocity (which is 0 since the ball is thrown horizontally)
g is the acceleration due to gravity (-9.8 m/s^2)
t is the time of flight
Rearranging the equation to solve for t, we get:
t = sqrt(2h/g)
Plugging in the given values, we have:
t = sqrt(2 * 50.0 / 9.8)
t = sqrt(102/9.8)
t ≈ 3.19 seconds (rounded to two decimal places)
Now, we can find the horizontal distance the ball covers during this time. We know the horizontal velocity is 10.0 m/s, and the time of flight is approximately 3.19 seconds.
d = vt
Where:
d is the horizontal distance
v is the horizontal velocity
t is the time of flight
Plugging in the values, we get:
d = 10.0 * 3.19
d ≈ 31.9 meters (rounded to one decimal place)
Therefore, the ball would hit the ground approximately 31.9 meters away from the base of the cliff.