A ball is thrown directly up in the air from a height of 5 feet with an initial velocity of 60 ft/s. Ignoring air resistance, how long until the ball hits the ground? Use the formula h=-16t^2+60t+5, where h is the height of the ball in feet and t is the time in seconds since it is thrown. Round your answer to the nearest tenth
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I say it's B not sure tho
'm gonna give you a short way to do this
Let us name our vectors! Anything going up is positive and anything coming down / acting down is negative.
u = +60 ft/s
g = -9.8 m/s^2 (since gravity acts downward) In this case though, it is m/s^2 and not ft/s^2....Change metres to feet and we will get -32.1522 ft/s^2
s = -5 ft (s is displacement which is vector and that is the direction the ball will be falling when coming down so we have taken into consideration direction).
t = what we are looking for
We can therefore use the equation:
s = ut + 0.5at^2
We can modify ti to get:
s = ut + 0.5gt^2
Plug in values:
-5 = (60)t + 0.5(-32.1522)t^2
-5 = 60t - 16.0761t^2
FORM INTO A QUADRATIC EXPRESSION:
-16.0761t^2 + 60t + 5 = 0
Solve for t by using the quadratic equation:
You get t = -0.08 s (1 d.p) and t = 3.8 s
TIME CAN NEVER BE NEGATIVE, so we will use the POSITIVE value for the time
Hence, the time for trajectory (movement of the ball) is 3.8 seconds