a. Assuming all energy involved is conserved, at the lowest point of the swing (which includes the moment the student grabs the rope), the student only has kinetic energy,
K = 1/2 m (3.5 m/s)²
and at the highest point of the swing, the student only has potential energy
P = mgh
The energies at the bottom and top of the swing must be equal, so
1/2 m (3.5 m/s)² = mgh
h = (3.5 m/s)² / (2g)
h = 0.625 m ≈ 63 cm
b. In part (a), we found the relationship
h = v²/(2g)
If we cut the speed v in half, we replace v in the equation above with v/2 :
h = (v/2)²/(2g)
and simplifying this gives
h = (v²/4)/(2g) = 1/4 • v²/(2g)
The factor of 1/4 tells you that reducing the speed by a factor of 1/2 reduces the height by a factor of 1/4. So he can swing as high as
1/4 (3.5 m/s)²/(2g) = 0.15625 m ≈ 16 cm
