Question

A ball is thrown straight upward from ground level with a velocity of 18m/s How much time passes before the ball strikes the ground? Disregard air resistance.

Answer 1

To determine the time before the ball strikes the ground, we can use the equation for vertical motion and solve for time. Using the given initial velocity of 18 m/s and disregarding air resistance, we find that it takes approximately 3.673 seconds for the ball to reach the ground.

Step-by-step explanation:

To determine how much time passes before the ball strikes the ground, we can use the equation for vertical motion:

h = ut + 0.5gt^2

Where h is the height, u is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. Since the ball is thrown straight upward, the initial velocity is positive (+18 m/s). The height is zero because the ball eventually reaches the ground. Rearranging the equation, we have:

0 = 18t - 0.5(9.8)t^2

Simplifying, we get:

4.9t^2 - 18t = 0

Factoring out a t, we have:

t(4.9t - 18) = 0

This equation has two solutions: t = 0 and t = 18/4.9 ≈ 3.673 seconds. Since time cannot be negative, the ball takes approximately 3.673 seconds to strike the ground.

Answer 2

The displacement of the ball can be calculated using:
s = ut + 1/2 at²; where u is the initial velocity and a is the acceleration. Because these are in opposite directions, one will be given a negative sign.
s = 0 at ground level
s = 18t - 1/2 x 9.81t²
t(18 - 4.9t) = 0
t = 0; t = 3.67 seconds
Thus, the ball will come back and strike the ground at 3.67 seconds.