SOLUTION: A ball is launched at 19.6 meters per second from a 58.8 meter tall platform. The equation for the object’s height h (in meters) at time t (in seconds) after launch is h(t)=-4.9t
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Question 1177623: A ball is launched at 19.6 meters per second from a 58.8 meter tall platform. The equation for the object’s height h (in meters) at time t (in seconds) after launch is h(t)=-4.9t^2+19.6t+58.8. How high does the ball fly in the air? When does the ball hit the ground?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
A ball is launched at 19.6 meters per second from a 58.8 meter tall platform. The equation for the object’s height h (in meters) at time t (in seconds) after launch is h(t)=-4.9t^2+19.6t+58.8.
h(t)=-4.9t^2+19.6t+58.8.
t= -b/2a = -19.6/2*(-4.9) =2
S(2) = 4.9(2^2) +19.6*2 +58.8
S(2) = 78.4 ft.
Maximum height = 78.4 ft
When it hits the ground height is 0
h(t)=-4.9t^2+19.6t+58.8.
-4.9t^2+19.6t+58.8 =0
/4.9
-t^2 +4t +12=0
(t-6)(t+2)=0
t=6 , it hits the ground
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