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)
Please help me with this problem, thanks.
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)
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.