A 4 kg projectile has 1400 J of kinetic energy when it leaves the ground. If it has 120 J of kinetic energy at its highest point, how high did it travel?

Answer 1

KE = 0.5*M*Vo^2 = 1400 J.
Vo^2 = 1400/0.5M = 1400/2 = 700
Vo = 26.5 m/s. = Initial velocity.

KE = 0.5*M*Xo^2 = 120 J.
Xo^2 = 120/0.5M = 120/2 = 60
Xo = 7.75 m/s=Hor. component of initial
velocity.

Xo = Vo*Cos A = 7.75 m/s.
26.5*Cos A = 7.75
Cos A = 0.29230
A = 73o

Yo = Vo*sin A = 26.5*sin 73 = 25.3=Ver.
component of initial velocity.

Y^2 = Yo^2 + 2g*h = 0
h = -(Yo^2)/2g = -(25.3^2)/-19.6=32.8 m.