A body is projected with velocity u at angle θ with vertical. At what height will it strike the wall, r distance away from the point of projection?
Consider a body which is projected from O in XY plane with velocity 'u' at angle θ with vertical, and The body is projected at an angle α = 90°–θ with the horizontal.
Let the body strike the wall at height h.
So, co-ordinates of point P where the projectile hits the wall are ↔ (r,h).
Equation of trajectory is given by,
[∵ ]
Since lies on trajectory,
∴ ,
is the height at which the body will strike the wall.
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A bullet fired from a gun with a velocity of 800 ft/s strikes the ground at the same level as the gun at a distance of 17320 ft. Find the angle of inclination with the horizontal at which the gun was fired.
Given that,
Distance of the gun, R = 17320 ft
Initial velocity, u = 800 ft./sec
Horizontal range is given by,
Therefore,
, is the angle of inclination with the horizontal at which the gun was fired.
A body is projected with kinetic energy 100J at an angle 45° with horizontal. What fraction of kinetic energy will be the potential energy at the highest point?
If a body is projected with kinetic energy K at angle θ, then kinetic energy at highest point is K cos2θ.
Therefore, potential energy of the body at highest point is,
P.E. =
Fractional potential energy at highest point is,
Here, we have
K = 100 J and Fractional potential energy at highest point is,
At the highest point, K.E will be 0.5 times the P.E at the highest point.
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Water stream leaves from hose pipe with velocity u m/s at angle θ with vertical and hose pipe rotates about vertical axis with angular velocity ω rad/sec. Find the area of circle and where will the water stream strike the ground?
The radius of the circle is equal to the horizontal distance at which water stream strikes the ground.
That is,
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A ball is projected at angle 0 from a point O ↔ (0,0). After passing the ball through a point (x, y) it reaches the ground at a distance R from O. Show that