Question 1100890: a projectile is fired vertically upward from the ground with a velocity of 38 m/s its distance above the ground is given
by s=-4.9t^2 + 38t where s is distance in m and t is the time in s,
a) determine how long it will take for the projectile to hit the ground.
b) determine how long it will take the project to reach a distance of 45 m above the ground.
Answer by ikleyn(52921)   (Show Source):
You can put this solution on YOUR website! .
a projectile is fired vertically upward from the ground with a velocity of 38 m/s its distance above the ground is given
by s=-4.9t^2 + 38t where s is distance in m and t is the time in s,
a) determine how long it will take for the projectile to hit the ground.
It will happen when s(t) = 0. To find "t", solve the equation
-4.9^t^2 + 38t = 0. ====>
One root is t = 0, bit it is not interesting to you, because it corresponds to the STARTING moment.
You are interested to find non-zero (positive) value of "t", when the projectile hits the ground.
Assuming that t=/= 0, you can divide both sides of this equation by "t".
You will get then -4.9t + 38 = 0 ====> 4.9t = 38 ====> t =  = 7.55 seconds (approximately).
So, the first question is answered.
b) determine how long it will take the project to reach a distance of 45 m above the ground.
To answer this question, you need to solve the equation s(t) = 45, which is
-4.9t^2 + 38t = 45.
Transform it to
4.9t^2 - 38 t + 45 = 0
and solve this quadratic equation using the quadratic formula.
You will get two roots:
one root will be solution for the ascending branch; the other root will be the solution for the descending branch.
Plot h(t) = y = -4.9*t^2 + 38t (red) and y = 45 (green)
Do you see the time moment t = 7.55 seconds, when the projectile hits the ground ?
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