Asked by Gerrad
If a ball is thrown upward from a building 30 m tall and the ball has a vertical velocity of 25 m/s, then its approximate height above the ground t seconds later is given by h(t) = 30 + 25t - 5t^2
a. After how many seconds does the ball hit the ground?
b. What is the domain of h?
c. How high does the ball go?
a. After how many seconds does the ball hit the ground?
b. What is the domain of h?
c. How high does the ball go?
Answers
Answered by
bobpursley
a. use the equation given, h(t)=0, so
0=30+25t-5t^2 solve for t.
h domain is zero to max height
max height will be when velocity is zero (at top)
V=25-10t but v is zero, solve for t at max height, then put that t in the original h(t) equation.
0=30+25t-5t^2 solve for t.
h domain is zero to max height
max height will be when velocity is zero (at top)
V=25-10t but v is zero, solve for t at max height, then put that t in the original h(t) equation.
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