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At the highest point of oblique projection, which of the following is correct? |
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Answer» Solution :This projectile motiontakes place when the intial velocity is not horizontal, but at some angle with the verticl , as shown in FIGURE (Oblique projectiom). Example, Water ejected out of a hose pipe held obliquely. Cannon fired in a battle ground, consinder an object thrown with intial velocity at an angle `theta` with the horizontal. then ` vecu=u_(x)hati+u_(y)hatJ` Where `u_(x)=u costheta` is th horizontal componant and `u_(y)=u sintheta`, the vertical component of velocity. since the accelereration due to GRAVITY is in he DIRECTION opposite to te directionopposite to the direction of vertical component `u_(y)` this component will gradually reduce to zero at the MAXIMUMHEIGHT of the projectile . At this MAXIMUM height the same garvitational force will oush the projectile to move downward and fall to the ground. There is no acceleration along the x directionthroughout the motion , So the horizonral component of the velocity `(u_(x)=u cos theta)` remains the same till the object reches the ground.  Hence after the time t, the velocity along horizontal motion ` v_(x)=u_(x)+a_(x)t=u_(x)=u costheta`  Thuis , `x=u cos theta. t or t =(x)/(u costheta)` Here `u_(y)=usin theta, a_(y)=-g` (acceleration due to gravity acts opposite to the motion). Thus `v_(y)=u sin theta-g t` The vertical distance travlled by the projectile in the same time t is Here , `s_(y)=y, u_(y)=u sin theta, a_(x)=-g` Substitute the value of t from equation (1) in equation (2), we have `y=u sin theta (x)/(u costheta)-(1)/(2g)(x^(2))/(u^(2) cos^(2) theta)` `y=xtantheta -(1)/(2g)(x^(2))/(u^(2) cos^(2) theta)` |
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