Question

Part 1

y = yo + Vo sine t - 1/2 gt? Vy = Vo sine - gt X = Xo + Vo coset Vy = Vo cose The equations above can be used for this and the following problems. is the angle above the horizontal. Ignore air resistance. A bullet is fired from a rifle pointed 45 degrees above horizontal. The bullet leaves the muzzle traveling 1400 m/s. How many seconds does it take the bullet to reach the high point of its trajectory? Answer:

Part 2

Part 3

Part 1

y = y₀ + v₀ sinθ t - 1/2 gt²

v_y = v₀ sinθ - gt

x = x₀ + v₀ cosθ t

v_x = v₀ cosθ

The equations above can be used for this and the following problems.
θ is the angle above the horizontal. Ignore air resistance.

A bullet is fired from a rifle pointed 45 degrees above horizontal. The bullet leaves the muzzle traveling 1400 m/s. How many seconds does it take the bullet to reach the high point of its trajectory?

Part 2

What is the bullet's height at the highest point?
Express the answer in km.

Part 3

How far away from where the gun is fired does the bullet land?
Answer in km.

Answer 1

Solution: (1) The expression for the time of flight of a projectile motion is 2v, sin (D) t= The time taken to reach maximum height is expressed as v sin (O) T mat g Substituting we get, (1400m/s) sin (45) TO 9.8m/s2 = 1013 man

(2) The maximum height reached in a projectile motion is given by the expression as follows: visine H= 2g Substituting we get, (1400m/s) sin? 45 H= (2)(9.8m/s) = 50000 m =50km (3) The expression for horizontal range of projectile is v; sin (20) R g Substituting we get, (1400m/s) sin 45 R= (9.8m/s) =141.4 km