Reply Sat 6 Jun, 2020 12:10 pm
A ball was launched which followed the path h=t^2+8t+22 where t is time in seconds and h is height in feet
a) Determine the height of the platform

b) Determine. Algebraically, when the ball will touch the ground

c) Determine algebraically, how high the ball reached
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Reply Sat 13 Jun, 2020 05:50 am
I don't think you have the equation correct....the more time that passes, the higher your ball is going to go. perhaps you mean -t squared, not t squared....if that's the case, at time 0, you are at 22 feet, time 1 you are at 29 feet, at time 3 you are at 37 feet ,by time 8, you are heading back down at 22 feet again and so on....So you are looking at a parabola where the ball goes up and then goes down...The platform is 22 feet (time = 0)....the ball will touch the ground when h=0 (solve for t).....how high will the ball go?...find the vertex of the parabola.......hope this helps
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Reply Sat 13 Jun, 2020 06:52 am
There's probably a negative sign in front of the t^2 term.

a) Find the time at t=0
b) Find the time t when the height is zero.
c) Use the formula for the max/min of a parabola, t=-b/2a to find the time, then put that time back into the equation to find the height.
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