A projectile is shot upwards at a 60degree angle with the ground and a speed of 200m/s?

Question

1) What is the vertical and horizontal components?

2) How long was it in the air?

3) What was its maximum height?

4) How far did the projectile go?

Answer 1

1) What is the vertical and horizontal components?

Muzzle velocity V = 200 mps; therefore vx = V cos(60) and vy = V sin(60).

2) How long was it in the air?

T = tup + tdown; tup = tdown = t; so vy = gt and t = vy/g where g = 9.81 m/sec^2 and vy is from 1). Thus T = 2t, the total time in the air.

3) What was its maximum height?

PE(at max height h) = mgh = 1/2 mvy^2 = KE(up at Muzzle); so that h = vy^2/2g and vy came from 1)

4) How far did the projectile go?

X = vx*T; where vx = constant found in 1) and T = 2t found in 2)

There you are, you can do the math.

Here's the physics. The verical kinetic energy at the muzzle is equal to the potential energy at maxium height h because, in climbing up, that KE is converted to potential energy PE. The kinetic energy from the horizontal velocity is not converted to potential energy, it remains fixed at the muzzle KE throughout the flight.

The time up is the same as the time down. This results because the projectile is accelerating at the same rate up as it is down. That is, the acceleration is g the acceleration due to gravity. So the total time is just twice the time up.

Since vx = constant, the distance the projectile travels is just the total time in the air T = 2t = 2(vy/g) times the fixed vx.