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?
- Audrey HepburnLv 71 decade agoFavorite Answer
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~~~