1.A ball is dropped from the height of 20m.calculate
(i) the time taken by the ball to reach the ground.
(ii) the velocity with which the ball strikes the ground.
hf=hi+vi*t-4.8t^2
0=20+0*t-4.8t^2
solve for t.
vf=vi+9.8t, using t from above, solve for vf
(i) To calculate the time taken by the ball to reach the ground, we can use the equation of motion:
h = (1/2) * g * t^2
Where:
h = initial height of the ball (20 m)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken by the ball to reach the ground (unknown)
Rearranging the equation, we have:
t^2 = (2h) / g
Substituting the given values, we get:
t^2 = (2 * 20) / 9.8
t^2 = 40 / 9.8
t^2 ≈ 4.08
Taking the square root of both sides, we find:
t ≈ √4.08
t ≈ 2.02 seconds (approximately)
Therefore, the ball takes approximately 2.02 seconds to reach the ground.
(ii) To calculate the velocity with which the ball strikes the ground, we can use another equation of motion:
v = g * t
Where:
v = final velocity of the ball (unknown)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken by the ball to reach the ground (2.02 seconds, as calculated in part (i))
Substituting the given values, we get:
v = 9.8 * 2.02
v ≈ 19.8 m/s (approximately)
Therefore, the velocity with which the ball strikes the ground is approximately 19.8 m/s.