A projectile is launched with an initial velocity of 25 m/s at a launch angle of 36

Question

degrees. What is the magnitude of the projectile’s velocity when it is 7.5 m above the ground?

Answer 1

v0=25 m/s
θ=36° (assumed with horizontal)
vh=v0 cos(θ)
vv=v0 sin(θ)

Let v=vertical velocity at any height S from ground, and g=9.81 m/s^2
Use v²-vv²=-2gS (for vertical direction)
v²=vv²-2gS
By kinematics:

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)

V=sqrt(v²+vh²)
=sqrt(vv²-2gS+vh²)
=sqrt(v0²-2gS)
=sqrt(25²-2*9.81*7.5)
=21.86 m/s
Note that the magnitude does not depend of the angle θ.

By energies:
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
=sqrt(2(kinetic energy)/m)
=sqrt(2((1/2)25^2-9.81*7.5))
=sqrt(477.85)
=21.859
as before.