# physics---projectile

1. A projectile is fired with an initial speed of 75.2m/s at an angle of 34.5degree

above the horizontal on a long flat firing range. What is the velocity of the

projectile 1.5s after firing?

Rating
• 1 decade ago

Vertical motion and horizontal motion are independent.Firstly, consider the horizontal motion, the projectile travels at constant horizontal speed.

Horizontal speed, vh = 75.2cos34.5* = 61.974 ms-1

Lastly, we consider the vertical motion, take upward direction be positive.

Initial speed, u = 75.2sin34.5 ms-1

Time of flight, t = 1.5 s

Acceleration due to gravity, g = -10 ms-2

By v = u + gt

v = 75.2sin34.5* + (-10)(1.5)

vv = 27.594 ms-1

Therefore, the resultant speed

= √(vh2 + vv2)

= √[61.9742 + 27.5942]

= 67.8 ms-1

Let x be the angle between the velocity and the horizontal

tanx = vv / vh = 27.594 / 61.974

x = 24.0*

So, the velocity of the projectile 1.5s after firing is 67.8 ms-1 in the direction 24.0* above the horizon.

Source(s): Myself~~~