1)A hiker throws a ball at an angle of 21.0 above the horizontal from a hill 21.0 m high. The hiker's height is 1.750 m. The magnitudes of the horizontal and vertical components are 14.004 m/s and 5.376 m/s, respectivley. Find the distance between the base of the hill and the point where the ball hits the ground. (Consider the hiker's height while calculating the answer.)
just te1l me what direction to go in please.
2)A car of mass 1330 kg is traveling at 28.0 m/s. The driver applies the brakes to bring the car to rest over a distance of 79.0 m. Calculate the retarding force acting on the car.
physics - tchrwill, Wednesday, April 16, 2008 at 1:53pm
Assume the ball starts its upward flight from the top of the throwers head, 1.75m above the ground.
1--Time to maximum height above the thrower's head derives from Vf = Vo - 9.8t or 0 = 5.376 - 9.8t making t(up) = .548sec.
2--Time back to the top of the throwers head is also .548sec. and the velocity the same as the departure velocity, 5.376m/s.
3--The time to reach the ground from the height of 21 + 1.75 = 22.75m = Vot + 9.8t^2/2 or 4.9t^2 + 5.376t - 22.75 = 0.
4--Solve for t using the quadratic formula.
5--Having t, the horizontal distance traveled becomes 14.004(.548 + .548 + t)
physics - Anonymous, Wednesday, April 16, 2008 at 2:14pm