as usual we take the rightward horizontal direction to be 0degrees and upward to

Question

as usual we take the rightward horizontal direction to be 0degrees and upward to be 90 degrees. A 1.5 kg basketball is thrownto the floor. At the momentit first hits the floor, the basketballhas a velocity of 6 m/s directed 45 degrees below the horizontal.In contact with the ball for 0.1 sec, the floor exerts upon theball a force of 135N directed 110 degrees above the horizontal.what is the velocity(magnitude and direction) of the ball at themoment it leaves the floor? Once the ball leaves the floor it is in projectile motion. howhigh will the ball bounce, and through what horizontal distancewill the ball move, before hitting the floor again? as usual we take the rightward horizontal direction to be 0degrees and upward to be 90 degrees. A 1.5 kg basketball is thrownto the floor. At the momentit first hits the floor, the basketballhas a velocity of 6 m/s directed 45 degrees below the horizontal.In contact with the ball for 0.1 sec, the floor exerts upon theball a force of 135N directed 110 degrees above the horizontal.what is the velocity(magnitude and direction) of the ball at themoment it leaves the floor? Once the ball leaves the floor it is in projectile motion. howhigh will the ball bounce, and through what horizontal distancewill the ball move, before hitting the floor again?

Explanation / Answer

since the force on the ball F is given           Fx cos(70) = m(vx-vx')/t m = mass of the ball vx = horizatal component of velocity of the ball     = vcos simillarly for the verticle component of force wecan solve for verticle compoent of velocity and the final velocity of the ball         V = (Vx^2+Vy^2)   

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