Projectile Motion - Help againWatch
(a) A cricket ball is thrown with a velocity of 31.4 ms-1 at an elevation angle
of 49 degrees.
Ignoring effects of air resistance, determine:
(i) The maximum height obtained.
(ii) The time of flight and the range on a horizontal plane.
(iii) The velocity (magnitude and direction) at a height of 18.0 m.
(b) Determine the velocity of throw required at an elevation angle of 41 degrees to reach the ball catcher at -1.5 m elevation (catcher 1.5 m lower than the thrower’s arm) located at a horizontal distance of 87 m.
Ok, for part (i) i used the equation H = v^2sin^2x / 2g, and got the answer to be 28.7m to 3.s.f
for part (ii) i used the equation t = 2vsinx / g, and got the answer to be 4.83s to 3.s.f
and for the horizontal range i used R = v^2sin2x / g which gave me 99.5m to 3.s.f
for part (iii) im having some difficulty with this i used the equation H = v^2sin^2x / 2g and made 'V' the subject to work out the velocity which gave me the answer to be 24.9m/s + and - and this is the speed twice - whilst going up and then coming back down again hence the signs however how do i workout the direction?
and for part B i still dont know how to do that question please can someone help me
You already have a thread for it where you have been geting help.
* Use v²=u² + 2as
where u is the vertical component of the initial velocity.
* v is the vertical component you are looking for
s in the vertical height reached
You then have two components of velocity.
The angle is found from trig using tangent.