the horizontal
displacement can be expressed as:
Δx=vxt
Height is expressed as Δy=vyt−12gt2
The time it would hit the maximum
height is expressed as:
0=vy−gt→t=vyg
The time that the projectile reaches
the ground is twice the time it takes the projectile to reach its maximum height.
This is expressed as
t=2vyg
to find the time of the projectile
to reach the maximum height, we have the formula
r=vx2vyg
From trigonometry,
vx=cos(θ)vy=sin(θ)
r=2cos(θ)sin(θ)g
We know from our trigonometric
identities that
sin(2θ)=2sin(θ)cos(θ)
, so it would become
r=v2gsin(2θ)
ymax=v2yg−12g(vyg)2=v2yg−12v2yg=v2sin2(θ)2g
velocity
is R=301.5
θ
=25 g=10
m/s2
V(0)=? Y(max)=?
R=R=v2∗sin(2θ)/g
Substitute the answers in the equation v ll, that would be V^2=3935m/s we also know that at the highest
point V(y)=0 so we v can write down:
V(y)2−V(0)(y)2=−2gΔy
we also know that
V(y)=V(0)∗sinθ
V(y)2−[V(0)sinθ]2=−2gΔy
y :35m
