Asked by Anonymous
A ball is thrown vertically upwards From the top of a building of height 29.4 m and with an initial velocity 24.5 m/sec. If the height H of the ball from the ground level is given by H = 29.4 + 24.5t - 4.9t², then find the time taken by the ball to reach the ground
Answers
Answered by
Henry
V = Vo + g*Tr = 0.
24.5 -9.8*Tr = 0,
Tr = 2.5 s. = Rise time or time to reach max height(H).
H = 29.4 + 24.5*2.5 - 4.9*2.5^2 = 60 m. Above gnd.
H = 0.5g*Tf^2 = 60.
4.9Tf^2 = 60,
Tf = 3.5 s. = Fall time.
T = Tr + Tf = 2.5 + 3.5 = 6 s. = Time in flight.
24.5 -9.8*Tr = 0,
Tr = 2.5 s. = Rise time or time to reach max height(H).
H = 29.4 + 24.5*2.5 - 4.9*2.5^2 = 60 m. Above gnd.
H = 0.5g*Tf^2 = 60.
4.9Tf^2 = 60,
Tf = 3.5 s. = Fall time.
T = Tr + Tf = 2.5 + 3.5 = 6 s. = Time in flight.
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