A child drops a ball from a height of 10 m. Assume its acceleration increases uniformly at the rate of 10m/s2. find the velocity with which the ball strikes the ground. Find the time taken to the ball reach the ground.
V^2 = Vo^2 + 2g*h = 0 + 20*10 = 200
V = 14.14 m/s.
0.5g*t^2 = 10 m.
5t^2 = 10
t^2 = 2
t = 1.414 s.
To find the velocity with which the ball strikes the ground and the time taken for it to reach the ground, we can use equations of motion. Let's break down the problem step by step.
Step 1: Finding the time taken (t) for the ball to reach the ground.
We can use the first equation of motion:
s = ut + (1/2)at^2
Where:
s = height (distance) covered by the ball = -10 m (negative because dropping downwards)
u = initial velocity = 0 m/s (since the ball was dropped)
a = acceleration = 10 m/s^2
t = time taken to reach the ground (which we need to find)
Plugging in the given values, the equation becomes:
-10 = 0 * t + (1/2) * 10 * t^2
-10 = 5t^2
Dividing both sides by 5:
-2 = t^2
Since time cannot be negative, we take the positive square root:
t = √2
Therefore, the time taken for the ball to reach the ground is approximately √2 seconds.
Step 2: Finding the velocity (v) with which the ball strikes the ground.
We can use the second equation of motion:
v = u + at
Where:
v = final velocity (velocity with which the ball strikes the ground)
u = initial velocity = 0 m/s (since the ball was dropped)
a = acceleration = 10 m/s^2 (positive because the ball is accelerating downwards)
t = time taken to reach the ground (which we found in the previous step, approximately √2 seconds)
Plugging in the given values, the equation becomes:
v = 0 + 10 * √2
v = 10√2 m/s
Therefore, the velocity with which the ball strikes the ground is approximately 10√2 m/s.