posted by Candice on .
Hi, could you please help with this question :)
Find the speed and direction of a particle which, when projected from a point 15 m above the horizontal ground, just clears the top of a wall 26.25 m high and 30 m away.
Thanks in advance
h1 = 15 m.
h2 = 26.25 - 15 = 11.25 m. = Ver. dist.
from firing point to top of wall.
Y = h2 = 11.25 m.
X = 30 m.
tanA = Y/X = 11.25 / 30 = 0.375
A = 20.6 Deg. = Direction of particle.
Range = Vo^2*sin(2A)/g = 30 m.
Vo^2*sin(41.2) / 9.8 = 30
Divide both sides by sin(41.2):
Vo^2 / 9.8 = 45.5
Vo^2 = 446.3
Vo = 21.1 m/s @ 20.6 Deg. = Velocity