Help plssss 30 point this time
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Refer to the diagram shown below.
We want to find V, the initial launch velocity of a projectile, without knowing θ, the launch angle.
To find V, we need to perform an experiment to determine:
d, the horizontal distance traveled,
t, the time taken to travel the horizontal distance.
Assume that aerodynamic resistance is negligible.
The horizontal component of the velocity is
Vx = V cosθ
Because the horizontal distance, d, is traveled in time, t, therefore
(V cosθ)*t = d
Vcosθ = d/t (1)
Assume that ground level has height zero.
Note that g, the acceleration due to gravity is known.
The vertical travel between the time of launch and return to ground level obeys the equation
Vsinθ*t - (g/2)*t² = 0
Therefore
t[Vsinθ - (gt)/2] = 0
Obtain
t = 0, which corresponds to launch
or
t = (2Vsinθ)/g
That is,
Vsinθ = (gt)/2 (2)
From (1) and (2), obtain
(Vcosθ)² + (Vsinθ)² = (d/t)² + (g²t²)/4
V²(cos² + sin² ) = (d/t)² + (g²t²)/4
Because g,t and d are known, V can be calculated.
We want to find V, the initial launch velocity of a projectile, without knowing θ, the launch angle.
To find V, we need to perform an experiment to determine:
d, the horizontal distance traveled,
t, the time taken to travel the horizontal distance.
Assume that aerodynamic resistance is negligible.
The horizontal component of the velocity is
Vx = V cosθ
Because the horizontal distance, d, is traveled in time, t, therefore
(V cosθ)*t = d
Vcosθ = d/t (1)
Assume that ground level has height zero.
Note that g, the acceleration due to gravity is known.
The vertical travel between the time of launch and return to ground level obeys the equation
Vsinθ*t - (g/2)*t² = 0
Therefore
t[Vsinθ - (gt)/2] = 0
Obtain
t = 0, which corresponds to launch
or
t = (2Vsinθ)/g
That is,
Vsinθ = (gt)/2 (2)
From (1) and (2), obtain
(Vcosθ)² + (Vsinθ)² = (d/t)² + (g²t²)/4
V²(cos² + sin² ) = (d/t)² + (g²t²)/4
Because g,t and d are known, V can be calculated.