A projectile is shot at an angle 45 degrees to the horizontalnear the surface of

Question

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.When it reaches the highest point of its trajectory, its speed is150 m/s. In a second trial with the same projectile, the initialspeed is the same but the angle is now 37 degree with thehorizontal. At its highest point in this trajectory, the velocityof the projectile would be what? I need help solving this problem. I got stuck what is theinitial velocity..need help A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.When it reaches the highest point of its trajectory, its speed is150 m/s. In a second trial with the same projectile, the initialspeed is the same but the angle is now 37 degree with thehorizontal. At its highest point in this trajectory, the velocityof the projectile would be what? I need help solving this problem. I got stuck what is theinitial velocity..need help

Explanation / Answer

in the first case let initial velocity of the projectile=v initial horizontal velocity of the projectile=v*cos45 horizontal acceleration=0 initial vertical velocity=v*sin45 vertical acceleration=g=9.8 m/s2 at the highest point the projectile's vertical velocity=0 at the highest point the projectile has only horizontalvelocity=v*sin45 v*cos45=150 v=212.16 m/s in the second case initial velocity=212.16 m/s initial horizontal velocity=212.16*cos37 horizontal velocity does not changes since the horizontalacceleration=0 at the highest point the projectile has only horizontal velocityvelocity=212.16*cos37=169.44 m/s in the second case the velocity at the highest point=169.44 m./s(ans)

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