A ball is thrown from the top of a 50-ft building with an upward velocity of 24 ft/s. When will it reach its maximum height? How far above the ground will it be? Use the equation h(t)= -16t2 + 24t + 50
Please help me with this problem, thanks.
g = 32 ft/s^2 approximately
v = Vi - g t
at top v = 0
0 = 24 - 32 t
solve for t
at that top point
h = Hi + Vi t - 16 t^2
h = 50 + 24 t - 16 t^2 (oh sorry, you know that already)
Thank you!
To find the maximum height, we need to determine the time it will take for the ball to reach its peak. To do this, we can use the equation for the time it takes for an object to reach its maximum height (t = -b / 2a).
In the given equation h(t) = -16t^2 + 24t + 50, we can see that a = -16 and b = 24. Plugging these values into the equation for finding the time to reach maximum height, we have:
t = -24 / (2 * -16)
t = -24 / -32
t = 0.75 seconds
So, it will take 0.75 seconds for the ball to reach its maximum height.
To find the maximum height, we can substitute the value of t = 0.75 seconds into the equation h(t) = -16t^2 + 24t + 50.
h(0.75) = -16(0.75)^2 + 24(0.75) + 50
h(0.75) = -16(0.5625) + 18 + 50
h(0.75) = -9 + 18 + 50
h(0.75) = 59 feet
Therefore, the ball will reach its maximum height at 0.75 seconds and it will be 59 feet above the ground at that point.