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### projectile motion

Posted: Wed Jul 16, 2014 8:31 pm
hey could i get help with this please?

A stone is projected at a cliff of height h with initial speed 42m/s at angle 60 degrees to horizontal. stone A lands after 5.5 secs at height 52 m. The max height reached is 68m. How do I find the speed of the stone just before impact?

### Re: projectile motion

Posted: Thu Jul 17, 2014 2:14 pm
Once you have the equations of motions for vertical and horizontal components of velocity and displacement in terms of t, substitute y = 52 and t = 5.5 into the equation for y(t), where y is the height in metres and t is the time in seconds. This will let you solve for h, since h appears in the equation for y(t).

Now that you know h, set y(t) = 0 and solve this quadratic in t to find the time T when the stone hits the ground (at y = 0).

By the Pythagorean Theorem, speed at any time will be equal to sqrt(ẋ²+ẏ²), where ẋ and ẏ are the horizontal and vertical components of velocity, respectively. Since you have the equations of ẋ and ẏ as functions of time, substitute t = T (which you found) into the equations for ẋ(t) and ẏ(t), and then substitute these values into sqrt(ẋ²+ẏ²) to get the speed at the time of hitting the ground.

I think the speed at the time of impact was referring to when y = 0. If it was referring to when y = 52, you know t = 5.5, so use that value of t to get speed = sqrt(ẋ²+ẏ²).

### Re: projectile motion

Posted: Fri Jul 18, 2014 10:10 am
Just adding to a.s.h.'s good explanation:

Once you have the time of impact, you don't need to use calculus to get the speed for this one. Neglecting air resistance,
v_x = constant = v0cos(launch angle)
v_y = v0sin(launch angle) - gt.
So, at any time t, just calculate v from sqrt(v_x^2 + v_y^2)

Joe