A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on a level surface. It travels in the air for 13.6 seconds before it strikes the ground 760 m from the shooter. At what angle above the horizontal was the bullet fired? Round to the nearest whole number and include units in your answer Use g= -9.8 m/s2 for the acceleration of gravity.

50°

Explanation:

Given the following :

Initial Velocity of bullet (U) = 87m/s

Time of travel (t) = 13.6s

Horizontal Distance traveled (S) = 760m

Therefore, the horizontal angle of projection of the bullet :

Using the second equation of motion:

S = ut + 0.5at^2

Where a = g = acceleration due to gravity, S = distance traveled, t= time taken and U = Initial Velocity.

The Angle of projection along the horizontal is represented as cosΘ

Acceleration due to gravity after the bullet has hit the ground = 0

Therefore, rewritten the equation :

S = ucosΘ * t + 0.5at^2

760 = 87 * Cosθ * 13.6 + 0.5(0)(13.6)^2

760 = 1183.2 * Cosθ + 0

Cosθ = 760 / 1183.2

Cosθ = 0.6423

Θ = cos^-1(0.6423)

Θ = 50.036460

Θ = 50°

real, smaller and upside-down.   so the answer should be b.

2b2t

explanation:

2b2t