Posted by **Cody** on Monday, February 6, 2012 at 3:07pm.

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?

- kinematics -
**MathMate**, Monday, February 6, 2012 at 4:10pm
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.

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