Kinematics?
You and a friend are throwing a tennis ball up against a 4.05 m building. You make a bet on whether or not you could throw the ball hard enough so you could get it on top of the building.
a) What would the initial speed of the tennis ball need to be in order for you to
throw it straight up and reach the height of the building?
b) Assuming the ball makes it up to
the top of building, how fast will the ball be traveling when it hits the ground after your
friend gets up on top of the building and drops the ball back down?
The answer for both is 8.91, but how do you get to that?
Answers
oldschool
Vyi²/19.6 = 4.05 Vyi = √(19.6*4.05) = 8.91m/s Vf² = 0²+2*g*d = 19.6*4.05 Vf = √(19.6*4.05) = 8.91m/s
az_lender
The downward acceleration of the ball will be 9.8 m/s^2, due to gravity. A suitable kinematic equation is 2ay = (vf)^2 - (v0)^2. In your problem a = -9.8 m/s^2, y = 4.05 m, and vf = 0 m/s. So (v0)^2 = 2*(9.8 m/s^2)*(4.05 m) = 79.38 m^2/s^2, and hence v0 = 8.91 m/s. The equation related to dropping the ball back down is exactly the same equation, except that the roles of (vf) and (v0) are reversed.
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