a ball is thrown directly upward with a velocity of 40 feet/seconds. its height in feet after t seconds is given by h(t)equals 40t-16t squared
To find the height of the ball at a given time, we can use the equation h(t) = 40t - 16t^2, where h(t) is the height in feet at time t.
Given that the ball is thrown directly upward with an initial velocity of 40 feet/second, it means that the initial height of the ball is 0 when t = 0.
To find the maximum height reached by the ball, we need to determine the time at which the ball reaches its peak. At the peak, the vertical velocity becomes 0. We can use the equation for velocity, v(t) = 40 - 32t, where v(t) is the velocity in feet/second at time t, and set it equal to 0:
40 - 32t = 0
Solving for t gives:
32t = 40
t = 40 / 32
t = 5/4
So, after 5/4 seconds, the ball reaches its maximum height.
To find this maximum height, we substitute t = 5/4 into the equation h(t) = 40t - 16t^2:
h(5/4) = 40(5/4) - 16(5/4)^2
= 1/4 (200 - 16(25/16))
= 25 feet
Therefore, the maximum height reached by the ball is 25 feet.
To find the height of the ball at any given time, substitute the value of t into the equation h(t) = 40t - 16t^2.
Note: When using the formula h(t) = 40t - 16t^2, please keep in mind that it assumes there are no external factors affecting the motion of the ball, such as air resistance.