A student standing on the ground throws a ball straight up. The ball leaves the

Question

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 11.0 when the hand is 2.00 above the ground.

How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

Please be detailed in your responce, I'm having trouble figuring out what theorem or formula to use. I was thinking x=1/2(V0+V)t but i dont know what the x is. and i dont know what the acceleration is, so someone please help me!

Explanation / Answer

Let the downward direction be positive. Initial Height of the ball = 2m Initial velocity of ball is = 11m/sec upward = -11m/sec Acceleration = g downwards = 9.8m/s^2 We need to find t See there is only one equation of motion that can solve this in 1 equation that is s = ut + .5at^2 s = 2-0 = 2 u = -11 a = 9.8 t= ? => 2 = -11*t + .5 *9.8*t^2 => 4.8t^2 -11t -2 =0 => t = 2.46and t = -0.169 So time = 2.46 sec ( reject -.169 negative)

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