ball is tossed straight up from a height of 5 ft above ground with a intial velocity of 4 ft/sec. height of the ball t seconds later is given by h(t)=-16t^2+40t+5. what is max height reached by the ball?
thanks
2006-11-26 10:59:47 · 6 answers · asked by matt b 2 in Education & Reference ➔ Homework Help
I'll solve this two ways, because I don't know whether you have taken calculus or not.
Since it's a quadratic, we can find the vertex (it's a max point). The value of x is -b/2a, where a is the t^2 coefficient and b is the t coefficient. -b/2a is the average of the roots, which is the x value of the vertex.
-b/2a = -40/-32=1.25 (this is the time)
Plug this into the equation to get 30. The max height it 30.
If you know calculus, you can use the derivative. If you don't know calculus, just skip this section.
f'(t) = -32t + 40
The zero will be the min/max in f(t), so
0=-32t+40
t=1.25
Plug it back in the original to get 30.
Hope this helps.
2006-11-26 11:10:32 · answer #1 · answered by Aegor R 4 · 0⤊ 0⤋
using calculus, dh/dt = 0 at h = maximum, and d^2h/dt^2 is negative at h max.
so differentiating, we get, dh/dt = 0= -32t + 40 and d^2h/dt^2 is negative.
so, t=1.25 seconds
substituting value of t in original equation ,we get,
max. height = -16(1.25^2) + (40*1.25) +5 = 30 feet.
******however you seem to have typed the initial velocity wrongly as 4 instead of 40...this is because your h equation is actually for vertical motion against gravity and is of the form
hf= ut - 1/2 g t^2 +ho
where,
hf = final height,
ho = initial height,
u= initial velocity,
g = acceleration due to gravity (32ft/sec^2),
t = time
2006-11-26 11:30:09 · answer #2 · answered by amateur_astrologer 2 · 0⤊ 0⤋
You don't NEED a graphing calculator for this. Basically you just solve for x,y. First, find the x (or t's in this case) when you find then via the quadratic equation, plug it back in and solve for y (or h(t) in this case) your max height will be the y in the points (x,y). I hope that helps. =)
2006-11-26 11:04:16 · answer #3 · answered by Seung Hee 5 · 0⤊ 0⤋
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2016-10-17 14:12:48 · answer #4 · answered by hultman 4 · 0⤊ 0⤋
put it in ur graphing calculator ( probably ti-83)
2nd calc> maximum > then put the range around the likely max
1.2500015
2006-11-26 11:05:09 · answer #5 · answered by AFizz 1 · 0⤊ 0⤋
you need a graphing calc.. ;)
i'd do this problem, since i love initial velocity problems, but my calculator is not handy... ha good luck though..
2006-11-26 11:01:46 · answer #6 · answered by jay 2 · 0⤊ 1⤋