How long until object hits the ground?
A projectile is fired upward from the ground with an initial velocity of 300 feet per second. Neglecting air resistance, the height of the projectile at any time t can be described by the polynomial function P(t) = ­16t2 + 300t How long will it be until the object hits the ground? I realize that my teacher...
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answers:
Steve4Physics say: h(t) = -16t² +300t+ h can’t be correct as you have h on both sides of the equation and h would cancel. But you are correct to put the minus sign in. h(t) = -16t² +300t.

When the object hits the ground h=0, so we just have to solve:
-16t² +300t = 0

75t – 4t² = 0 (divided-through by 4 and rearranged)
t(75 – 4t) = 0 (factorised)

t = 0 or t = 75/4 = 18.75s (=19s to 2 sig. figs.)

t = 0 corresponds to the time the projectile is fired; t = 19s is the time it returns to the ground.
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electron1 say: You are correct. The projectile’s acceleration is negative. The following equation is used to solve this problem.

ht = -½ * g * t^2 + 300 * t
½ * g = 16
g = 32 ft/s^2

a = -32 ft/s

During one half of the total time, the projectile’s velocity decreases from 300 m/s to 0 m/s. Let’s use the following equation to determine the time for this to happen.

vf = vi + a * t
0 = 300 + -32 * t
t = 300 ÷ 32 = 9.375 seconds
Total time = 18.75 seconds

I hope this is helpful for you.
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say: At 300 ft/sec and a downward acceleration due to gravity of 32.2 ft/sec^2, it takes 300 ft/sec / 32.2 ft/sec^2 = 9.32 sec to reach the top of the arc where its velocity = 0. The trip back takes the same amount of time.

2 * 9.32 sec is the total time.
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