A projectile is launched off of a 17.5 m high building with an initial velocity of 25.8 m/s at an angle of 31.6 degrees relative to the horizontal. How long is it in the air for?
Height of the building = 17.5 m, let it be the zero of our calculation (to measure the vertical distance).
Initial Velocity = 25.8 m/s at an angle of 31.6o relative to the horizontal.
Horizontal component of the velocity = 25.8 Cos( 31.6o), horizontal acceleration 0 m/s2
Vertical component of the velocity = 25.8 Sin( 31.6o), vertical acceleration = -g m/s2 = -10 m/s2 (say)
How long does the projectile stays in air ? It will stay i air till the vertical distance becomes -17.5 m
We are not concerned with the horizontal motion as it would just move the projectile in a uniform motion in the horizontal direction, still its hits the ground.
Vertical distance to be covered = -17.5 m
Vertical velocity = 25.8 Sin( 31.6o) = 13.51884 m/s
Vertical acceleration = -10 m/s2
Negative time solution is not physical
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