Question

7. Your friend's Frisbee has become stuck 27 mabove the ground in a tree. You want to dislodge the Frisbee by throwing a rock at it. The Frisbee is stuck pretty tight, so you figure the rock needs to be traveling at least 5.1 m/s when it hits the Frisbee.

If you release the rock 1.7 m above the ground, with what minimum speed must you throw it?

Answer 1

given that

Vf = 5.1 m/s

s = 27m - 1.7 m = 25.3 m

according to second law of motion

Vf^2 = Vi^2 + 2*a*s

where a = -g = -9.8 m/s^2

put the values in above equations

(5.1)^2 = Vi^2 + 2*(-9.8)*(25.3)

Vi^2 = 521.89

Vi = 22.84 m/s