posted by Cody on .
A projectile is launched with an initial velocity of 25 m/s at a launch angle of 36
degrees. What is the magnitude of the projectile’s velocity when it is 7.5 m above the ground?
θ=36° (assumed with horizontal)
Let v=vertical velocity at any height S from ground, and g=9.81 m/s^2
Use v²-vv²=-2gS (for vertical direction)
Magnitude of Velocity,V, is the vector
sum of the vertical and horizontal velocities, sqrt(v²+vh²).
Note that vh is constant with time, and
vv²+vh²=v0 (25 m/s)
Note that the magnitude does not depend of the angle θ.
Initial kinetic energy = (1/2)m(v0²)
Initial potential energy = 0 (assumed)
Final potential energy = mgH (H=7.5m)
Final Kinetic energy = (1/2)mv0²-mgH
Final magnitude of velocity