Plz help me with kinematics?

Question

a projectile is launched at 35 degrees above the horizontal with an initial velocity of 120m/s. What is the projectile's speed 3.00s later?

Answer 1

Initial Velocity, U= 120m/s

Horizontal Velocity always remains the same.

Hence, Initial Horizontal Velocity= Final Horizontal Velocity = 120 cos 35

=98.29824... m/s

Initial Vertical Velocity= 120sin35 = 68.82917...m/s

Using the formula  v= u + at

Taking a= -9.81 m/s^2

Final Vertical Velocity= 68.82917... + (-9.81 x 3)

                                 = 39.3991723... m/s

Using Pythagoras' Theorem,

Projectile's speed 3s later=  Ã¢ÂˆÂš (98.29824...^2) +(39.3991723..^2)

                                         = 105.900

                                         = 106 m/s (to 3.s.f)

Hope that helps =)