Let's write the equations of motion of the body in horizontal and vertical directions:
x=v0tcosθ,(1)y=v0tsinθ−21gt2,(2)
here, x is the horizontal displacement of the body (or the range), v0=24.5 sm is the initial velocity of the body, t is the time of flight of the body, θ=30∘ is the launch angle, y is the vertical displacement of the body (or the height) and g=9.8 s2m is the acceleration due to gravity.
a) Let's first find the time that the body takes to reach the maximum height from the kinematic equation:
vy=v0sinθ−gtrise,0=v0sinθ−gtrise,trise=gv0sinθ.
Then, we can find the time of flight of the body:
t=2trise=g2v0sinθ,t=9.8 s2m2⋅24.5 sm⋅sin30∘=2.5 s.b) We can substitute the time of flight of the body into the first equation and find the range of the body:
x=24.5 sm⋅2.5 s⋅cos30∘=53 m.c) Finally, we can substitute trise into the second equation and find the maximum height:
ymax=v0sinθ⋅gv0sinθ−21g(gv0sinθ)2,ymax=2gv02sin2θ,ymax=2⋅9.8 s2m(24.5 sm)2⋅sin230∘=7.66 m.Answer:
a) t=2.5 s.
b) x=53 m.
c) ymax=7.66 m.
Comments
Leave a comment