Aprojectile is launched in the horizontal direction. it travels 2.050 m horizontally while it falls 0.450 m vertically, and it then strikes the floor. a. how long is the projectile in the air? b. what was the original velocity of the projectile? c. what is the velocity in the horizontal direction of the projectile when it strikes the floor? d. what is its velocity in the vertical direction at this time? e. what is the magnitude of its velocity as it strikes the floor?

a) 0.303 s b) 6.77 m/s c) - 2.97  m/s d) 7.39 m/s

Explanation:

horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m

Using equation of linear motion

Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative

-0.45 = 0 - 0.5 × 9.81×t²

0.45 / (0.5 × 9.81) = t²

t = √0.0917 = 0.303 s

b) horizontal distance traveled = horizontal velocity (Ux) × t

Ux = horizontal distance traveled / t = 2.050 / 0.303 = 6.77 m/s

c) Velocity in the vertical direction can be calculated using

Vy = Uy - gt where g is negative since initial U is zero

Vy = - 9.81 × 0.303 = - 2.97  m/s

d) the magnitude of the velocity = resultant of the the two velocity

using Pythagoras theorem

the magnitude of the velocity = √ ( 6.77² + 2.97²) = 7.39 m/s

at angle = tan^-1 ( 2.97 / 6.77) =  23.69⁰

15.3 cos 65 degrees trig