a particle is projected with horizontal velocity u from the top of a plane at an angle theta with the horizontal please find the distance at which the particle strikes the plane

Question

anonymous · Answer

$$the \angle \theta $$ is the angle between the plane and the horizontal

anonymous · Answer

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anonymous · Answer

The x and y components are independent. Ignore x for now.

$$s=ut+0.5at^2  $$
$$a=-mg  $$
$$u=v_y(0)=v_0sin(\theta)  $$
$$s_y=ut+0.5at^2=v_0sin(\theta)t-0.5mgt^2 $$
At hitting the ground again,s_y=0 (evidently)
Rearrange to find the time
Insert the time into a similar equation for s_x, which I trust you can construct by yourself (or by analogy with the s_y one).

anonymous · Answer

Note: the equation for t you will get is a quadratic, so you will obtain 2 values for t. This, again, is obvious as there are 2 moments in time when the ball touches the floor (i.e. s_y=0), when it is launched and when it lands.

anonymous · Answer

$$0 = Vy*t _{f} -\frac{ g*t _{f}^{2} }{ 2 }$$
$$Vy =\frac{ g*t _{f}^{} }{ 2 }$$
$$ \frac{2V _{y} }{g} =t _{f}$$
$$\tan \theta =\frac{ V _{y} }{ u }$$
$$distance = u*t _{f}$$

anonymous · Answer

Sorry, $$F=mg=ma$$

anonymous · Answer

How amateurish of me.

anonymous · Answer

:)

anonymous · Answer

Although not intuitively obvious, with air resistance existing and all.

anonymous · Answer

air resistance is ignored until fluid dynamics