Motion Help! (1 Viewer)

[dishab]

A ball is dropped from a lookout 180 meters high. At same time, a stone is fired vertically upwards from the valley floor with speed V m/s. Take g = 10m/s^2.

a) Find for what values of V a collision in the air will occur. Find, in terms of V, the time and the height when collision occurs, and prove that the collision speed is V m/s.
b) Find the value of V for which they collide halfway up the cliff, and the time taken.

The answer for part a) is, for V greater or equal to 30m/s, they collide after 180/V seconds, 180/V^2 (V^2-900) meters above the valley floor.
b) V=30 sqaure root 2, 3 square root 2 seconds.

Any idea how to do this?

 

[MetroMattums]

For a), determine the equations for displacement against time for the two particles and the equation for velocity against displacement, then add them together and equate the new equation to 180 metres.

For b), substitute 90 metres into your equations.

Hopefully that should work :/

 

[fullonoob]

a)
rock speed :10t (downwards)
stone speed: v - 10t (upwards)

xr = -5t^2 + C
t = 0, xr =180
hence C = 180
xr = 180 - 5t^2 ->1

xs = vt - 5t^2 + D
t=0, xs=0
xs = vt - 5t^2 ->2

for collision xs = xr
vt - 5t^2 = 180- 5t^2
vt = 180
t = 180 / v seconds

xr = 0 at ground
0 = 180 - 5t^2
t^2 = 36
t = 6 , t>0
But, t<= 6 for collision to occur
hence v => 30 m /s for collision to occur
xr = 180 - 5(180/v)^2
= 180/ v^2 (v^2-900) from ground

b)
Half way : x = 90
90 = 180/ v^2 (v^2 - 900)
v^2 / 2 = v^2 - 900
v^2 = 1800
v = 30root 2 m/s , v>0
t = 180/v
= 180/ 30root2
= 3root 25

no im not using latex