Question

If a ball is thrown in the air with an initial

height of 5 feet, and if the ball remains in
the air for 4 seconds, then accurate to
the nearest foot, how high did it go?
Remember, the acceleration due to
gravity on Earth is -32 ft/sec2².
[?] feet

Answer 1

The height of the ball at any given time can be determined by the equation:

h(t) = -16t^2 + vt + h

where h(t) is the height of the ball at time t, v is the initial velocity of the ball (in feet per second), and h is the initial height of the ball (in feet).

In this problem, we know that the initial height of the ball is 5 feet, the time the ball remains in the air is 4 seconds, and the acceleration due to gravity is -32 ft/sec^2. We need to find the initial velocity of the ball (v) in order to determine the maximum height reached.

At the highest point of its trajectory, the velocity of the ball is zero. So we can use the equation:

0 = -32(4)^2 + v(4) + 5

Solving for v, we get:

v = 64 ft/sec

Now we can use this value of v to determine the maximum height reached by the ball:

h_max = -16(2)^2 + (64)(2) + 5

h_max = 69 feet

Therefore, accurate to the nearest foot, the ball went up to 69 feet