The ball is thrown vertically upwards from a height of 15 m above the ground. The initial speed of the ball is 10 m / s. How long does it take for the ball to hit the ground after being thrown? What is the speed of the ball just before hitting the ground?
H = 15 meters – the height from which the ball was thrown vertically upwards;
V0 = 10 m / s – initial ball velocity;
g = 10 m / s ^ 2 – acceleration of gravity.
It is required to find the time of the ball falling to the ground t (seconds) and the ball velocity V (m / s) at the moment of impact.
Since at the maximum point the speed of the ball will be zero, we find the flight time of the ball from the height H to the maximum height:
t1 = V0 / g = 10/10 = 1 second.
Then the ball will rise to a height:
H = H0 + V0 * t1 – g * t1 ^ 2/2 = 15 + 10 * 1 – 10 * 1/2 = 15 + 10 – 5 = 20 meters.
From this height, the ball will fall in the time:
t2 = (2 * H / g) ^ 0.5 = (2 * 20/10) ^ 0.5 = 4 ^ 0.5 = 2 seconds.
The total time of the ball falling to the ground from the moment of throwing will be equal to:
t = t1 + t2 = 1 + 2 = 3 seconds.
The speed of the ball at the moment of hitting the ground will be equal to:
V = g * t2 = 10 * 2 = 20 m / s.
Answer: the ball will fall to the ground 3 seconds after the throw, its velocity at the moment of impact will be equal to 20 m / s.