Asked by Mariam
A tennis ball is tossed up off a building with a velocity of 22m/s. It takes 6.4s to reach the ground. How high is the building and what is the maximum height of the tennis ball?
Answers
Answered by
bobpursley
hf=hi+vi*t-4.9t^2
so you know hf(=0), vi, time t, solve for hi. Looks straightforward to me.
so you know hf(=0), vi, time t, solve for hi. Looks straightforward to me.
Answered by
Henry
V = Vo + g*Tr = 0.
22 - 9.8Tr = 0.
Tr = 2.24 s. = Rise time.
Tr+Tf = 6.4.
2.24 + Tf = 6.4
Tf = 4.16 s. = Fall time.
h = 0.5g*Tf^2 = 4.9*4.16^2^2 = 84.8 m. above gnd. = ht. of tennis ball.
h = ho + Vo*Tr + 0.5g*Tr^2 = 84.8 m.
ho + 22*2.24 - 4.9*(2.24)^2 = 84.8.
ho + 24.69 = 84.8
ho = 60.1 m. = Ht. of the bldg.
22 - 9.8Tr = 0.
Tr = 2.24 s. = Rise time.
Tr+Tf = 6.4.
2.24 + Tf = 6.4
Tf = 4.16 s. = Fall time.
h = 0.5g*Tf^2 = 4.9*4.16^2^2 = 84.8 m. above gnd. = ht. of tennis ball.
h = ho + Vo*Tr + 0.5g*Tr^2 = 84.8 m.
ho + 22*2.24 - 4.9*(2.24)^2 = 84.8.
ho + 24.69 = 84.8
ho = 60.1 m. = Ht. of the bldg.
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