A projectale is launched from ground level with an initial speed of 53.2 m/s. Fi

Question

A projectale is launched from ground level with an initial speed of 53.2 m/s. Find the launch angle (the angle the initial velocity vector is above the horizontal) such that the maximum height of the projectile is equal to its horizontal range. (Ignore any effects due to are resistance-) A bal launched from ground level lands 2.2 s later on a level field 43 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.) At 1/2 of its maximum height, the speed of a projectle is 3/4 of its initial speed, what was its launch angle? (Ignore any effects due to air resistance.)

Explanation / Answer

21) From range equation R = u2 * sin (2 theta) / g

max height h = u2 * sin2(theta) / 2 g

equating the above equations

u2 * sin (2 theta) / g = u2 * sin2(theta) / 2 g

sin (2 theta) = sin2 (theta) / 2

2 sin(theta) * cos(theta) = sin2 (theta) / 2

tan theta = 4

theta = tan-1 (4)

launch angle = 75.96 deg

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