particle is projected from a point O on a horizontal plane with speed 35m/s at an angle of elevation α , where tan α =2. Find the maximum height above the plane of the particle and the equation of its trajectory.

Question

jamesj · Answer

Let Vx be the velocity of the particle in the horizontal direction
Vy the velocity of the particle in the vertical direction.

Then as the only force acting on the particle, gravity, is in the y-direction, then Vx is constant:

$$ V_x = V.\cos \alpha $$
where V is the initial velocity

On the other hand,
$$ V_y = V.\sin\alpha - gt $$ 

The projectile will be at its maximum height when $ V_y = 0 $.  Find that time t from the last equation.  Then substitute that into the equation for the y-location
$$ y(t) = V.\sin\alpha.t - \frac{1}{2}gt^2 $$

When you're back, let someone know if all this makes sense and I or someone else will help you with the trajectory.

anonymous · Answer

james is right equation of trajectory define from discard  time between these two equation :$$x(t)=(V _{0}\cos \alpha)t$$& $$y(t)=-(1/2)g  t ^{2}+(V _{0}\sin \alpha)t$$so yields:$$y=- (1/2V _{0}^{2}\cos ^{2}\alpha)g x ^{2}+x \tan \alpha$$