Students roll a snowball from a roof that makes an angle θ with respect to the horizontal. The snowball leaves the roof with speed V, at height H above the ground. A physics professor is standing a horizontal distance D away from the building; the professors head is a height h above the ground. Derive an expression for the initial speed V the snowball must have in order to hit the professors head. Then derive an expression for the speed with which the snowball actually hits the professor's head.

The spring constant K is computed with the information known about the mass at rest: 

F = kx = m*g = k*.2 

k = m*g/.2 = 6.8*9.81/.2 = 333.5 N/m 

we agree on K! 

The frequency of oscillation is: 

f = sqrt( k/m ) / ( 2*π ) = sqrt( 333.5 / 6.8 ) / ( 2*π ) = 1.11 Hz 

We agree here also! 

The kinetic energy at t = 0 is: 

E = (1/2)*m*v^2 = (1/2)*6.8*(4.6)^2 = 71.9 J 

At the extreme of motion, this translates entirely into additional spring potential energy. This point also represents the maximum acceleration. 

Ep = (1/2)*k*(Δx)^2 = E 

Δx = sqrt( 2*E / k ) = sqrt( 2*71.9 / 333.5 ) = .66 m 

The additional force of the spring is: 

F = k*Δx = 333.5*.66 = 219 N 

F = m*a 

a = F/m = 219/6.8 = 32 m/s^2 

a is the acceleration at maximum displacement, which is the maximum acceleration of the block, and so this is the answer to the second question. 

The equation of motion of the block is then: 

x = .2 + .66*Sin( 2*π*1.11*t) 

Choose the Sin term for the motion, since the additional displacement is zero at t = 0. 

The speed of the block is: 

v(t) = dx/dt =.66*[ Cos( 2*π*1.11*t ) ]*(2*π*1.11) 

v(.31) = .66*7.00*Cos( 7.00*.31 ) = -2.6 m/s 

This means that the mass is moving upward at 2.6 m/s. Note that the argument of the Cos is in radians. 

According to the equation of motion, the x displacement at 0.31 s is: 

x(.31) = 0.2 + .66*Sin( 7.00*.31 ) = 0.742 m 

This causes a spring force of: 

F = k*x = 333.5* ( .742 ) = 247 N