a. V = Vo + g*Tr = 0 @ max ht.
8 - 9.8Tr = 0
Tr = 0.82 s. = Rise time or time to reach max ht.
h = ho + Vo*Tr + 0.5g*Tr^2.
h = 2 + 8*0.82- 4.9*0.82^2 = 5.27 m.
above gnd.
h = 0.5g*Tf^2 = 5.27.
4.9(Tf)^2 = 5.27
Tf = 1.04 s. = Fall time.
Tr+Tf = 0.82 + 1.04 = 1.86 s. = Time in flight.
b. V^2 = Vo^2 + 2g*h.
Vo = 0, g = +9.8 m/s^2, h = 5.27 m., V = ?.
A basketball is thrown at 8.00m/s [up] from a height of 2.00m and eventually comes back down to hit the floor. a)assuming a uniform acceleration of 9.80m/s^2 [down], how long was the ball in flight? b) calculate the balls velocity immediately prior to hitting the floor? I need to use one of the five key equations for motion with uniform acceleration.
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