Asked by Anonymous
A projectile is shot at an angle of 40.0 degrees with a velocity of 25.0 m/s. If a building is 50.0 m away, will this projectile strike the building, and if so, at what height?
Answers
Answered by
drwls
The time to reach the building, if it does not hit the ground first, is
L/(Vo*cos40) = 50/(25 cos40) = 2.61 seconds
The time to hit the ground, if it does not hit the building first, is
2 Vo sin40/g = 3.28 s
Therefore it hits the building (assuming it is tall enough so that the projectile can't go over the top).
To find out where it hits, use the
y vs t equation:
y = Vo*sin40*t - (g/2)t^2
and use t = 2.61 s
L/(Vo*cos40) = 50/(25 cos40) = 2.61 seconds
The time to hit the ground, if it does not hit the building first, is
2 Vo sin40/g = 3.28 s
Therefore it hits the building (assuming it is tall enough so that the projectile can't go over the top).
To find out where it hits, use the
y vs t equation:
y = Vo*sin40*t - (g/2)t^2
and use t = 2.61 s
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