A projectile is fired from the ground (you can assume the initialheight is the s

Question

A projectile is fired from the ground (you can assume the initialheight is the same as the ground) in a field so there are noobstacles in its way. It is fired at an angle 34° with respectto the horizontal and with an initial speed of 47 m/s. Airresistance is negligible in this situation.The same projectile wasthen fired in the same way toward a wall that is a horizontaldistance 164.9 m from where the projectile was fired.

What was thevelocity of the projectile when it hit the wall?
Calldown the positive y direction,and toward the wall the positive x direction.
i +  j;

Explanation / Answer

   Given that initial speed of the projectile u = 47 m/s angle = 34o horizontal distance x = 164.9 m -------------------------------------------------------------------------    We know that            x= (u cos)t    Time t = x/(u cos) but vx = u cos         vy = usin - gt           The velocity of the projectile when it hitthe wall is             v = [vx2 +vy2] Substitute the values for v. horizontal distance x = 164.9 m -------------------------------------------------------------------------    We know that            x= (u cos)t    Time t = x/(u cos) but vx = u cos         vy = usin - gt           The velocity of the projectile when it hitthe wall is             v = [vx2 +vy2] Substitute the values for v.

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