Question

A ball is dropped from the top of a 550 ft. building. The function h\left( t \right) = - 16{t^2} + 550h(t)=−16t2+550 models the height of the ball, h(t) (in feet), at any given time, t (in seconds). How long does it take the ball to reach the ground?

Answer 1

To find the time it takes for a ball to reach the ground when dropped from a 550 ft building, the equation h(t) = -16t^2 + 550 is set to zero and solved for t. Discarding the negative root, the calculation shows that it takes approximately 5.86 seconds for the ball to reach the ground.

Step-by-step explanation:

The student is asking how long it will take for a ball dropped from the top of a 550 ft building to reach the ground. The motion of the ball is governed by the equation h(t) = -16t^2 + 550, where h(t) is the height of the ball above the ground at time t, in seconds. To find the time it takes for the ball to hit the ground, we need to find the value of t when h(t) = 0.

We set the equation to zero and solve for t:
-16t^2 + 550 = 0.
Factoring out the common term, we get:
-16(t^2 - 34.375) = 0.
Solving for t using the square root, we find that:
t^2 = 34.375
t = sqrt(34.375).
Because time cannot be negative, we discard the negative root and calculate the positive root:

t = 5.86 seconds (rounded to two decimal places).

Therefore, it will take approximately 5.86 seconds for the ball to reach the ground when dropped from the top of a 550 ft building, ignoring air resistance.

Answer 2

It has to be 3.64 because of the hieght