Question

A ball is thrown from the top of a 6m rock. Its height relative to the ground can be mapped by the following equatio -St^2+ 13t+ 6 = h(t). where t = time and h(t) = height above ground. Calculate the time at which the ball will hit the ground​

Answer 1

The ball hits the ground after 6 seconds.

Explanation:

To calculate the time at which the ball will hit the ground, we need to find the value of t when the height h(t) is equal to 0.

The equation for the height of the ball relative to the ground is:

h(t) = -St^2 + 13t + 6

We set h(t) to 0 and solve for t:

0 = -St^2 + 13t + 6

Rearranging and factoring, we get:

0 = (t - 6)(-St - 1)

So either t - 6 = 0 or -St - 1 = 0.

If t - 6 = 0, then t = 6, which means the ball hits the ground after 6 seconds.

If -St - 1 = 0, then t = -1/S, where S is the constant in the equation for the height of the ball. However, this solution is not physically meaningful because it implies that the ball would have been thrown from the ground with an initial velocity of -1/S, which is impossible.

Therefore, the ball hits the ground after 6 seconds.