Question 3274: SORRY! I caught a typo error in my question!! Given: h=-16 t squared +vt+ s, a ball can be thrown at 40 ft/s. Set the initial height at 0.
1. a.) When will the ball reach a height of 24 feet?
b.) When does the ball hit the ground?
2.) a.) If the ball is thrown at 64 ft/s, when will it reach 48 feet?
b.) When does the ball hit the ground?
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! h=-16 t^2 +vt+ s, s = 0,
1.a) v = 40 ft/s
When h = 24, solve the equation 24 = -16 t^2 + 40t or
2t^2 - 5t + 3 = 0,
Factoring (2t -3)(t - 1) = 0, so t = 1 or 3/2
Hence, when t = 1 or 1.5 sec, h = 24 ft
b) The height of the ground is h = 0, solve
-16 t^2 + 40t = 0 or 8t(2t -5) = 0.
So, t = 5/2 or 0(means initial)
Hence, when t = 2.5 sec, it hits the ground.
2.a) v = 64 ft/s
When h = 48, solve the equation 48 = -16 t^2 + 64t or
t^2 - 4t + 3 = 0,
Factoring (t -3)(t - 1) = 0, so t = 1 or 3
Hence, when t = 1 or 3 sec, h = 48 ft
b) The height of the ground is h = 0, solve
-16 t^2 + 64t = 0 or 16t(t -4) = 0.
So, t = 4 or 0(means initial)
Hence, when t = 4 sec, it hits the ground.
Kenny
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