A plane,diving at an angle of 37 $\circ$ below the horizontal,releases a projectile at an altitude of 730.0 m.The projectile hits the ground 5.00 s after being released.At what angle does the projectile strike the ground?
A)45.0 $\circ$ below the horizontal
B)43.4 $\circ$ below the horizontal
C)53.0 $\circ$ below the horizontal
D)37.0 $\circ$ below the horizontal
E)44.1 $\circ$ below the horizontal

Question

Quizplus · Accepted Answer

The projectile's vertical velocity is:$$v_{yf}=v_{yi}-gt=(730	ext{ m/s})\sin37\degree-(9.8	ext{ m/s}^2)(5.00	ext{ s})=-123	ext{ m/s}$$The projectile's horizontal velocity is:$$v_{xf}=v_{xi}=(730	ext{ m/s})\cos37\degree=588	ext{ m/s}$$The projectile's angle of impact is:$$	an	heta=\frac{v_{yf}}{v_{xf}}=\frac{-123	ext{ m/s}}{588	ext{ m/s}}=-0.209$$$$	heta=	an^{-1}(-0.209)=-11.9\degree$$The angle below the horizontal is:$$90\degree+11.9\degree=\boxed{44.1\degree}$$

A Plane,diving at an Angle of 37 ∘\circ∘ Below the Horizontal,releases a Projectile at an Altitude of the Horizontal,releases

A Plane,diving at an Angle of 37^$\circ$ Below the Horizontal,releases a Projectile at an Altitude of the Horizontal,releases