easy questions

[sinkokuyou]

1. Erica stands 20 ft away from a 10 ft tall wall. She throws a ball at a release height of 4 ft that just clears the wall. What is the ball release velocity?

2. Sand falls onto a conical pile at a rate of [unparseable or potentially dangerous latex formula]. The angle of resitution is the angle the pile makes with the ground and is [unparseable or potentially dangerous latex formula]. At what pile height is the pile radius increasing at [unparseable or potentially dangerous latex formula]?

[calcguy]

09H-37. The phrase "just clears the wall" tells you that the ball goes over the wall at it's maximum height (6 ft vertically from release point) which is half way horizontally from it's release point, 20 ft horizontally and 4 ft above the ground, to the point 20 ft horizontally on the other side of the wall and 4 ft from the ground.
Use tan(theta) = 4 hmax/Range to get the angle: tan(theta) = 4*(6 ft)/40 ft which gives theta = 31.0º
Now use Range = v^2*sin(2*theta)/g and solve for v: 40 = v^2*sin(2*31.0º)/32.17 and you get v = 38.2 ft/sec

09H-57. Given: dV/dt = 2 ft^3/min and dR/dt = 1 ft/min. Since the angle of restitution is 35º, the height of the cone is Rtan35º.
Cone volume is V = 1/3 pi R^3 tan35º so dV/dt = pi R^2 tan35º dR/dt = pi R^2 tan35º (1) = 2
Solve to get R = 0.954 ft so pile height = 0.954 tan35º = 0.668 ft.