Question

How to find initial velocity in projectile motion without angle?

Answer 1

To find initial velocity in projectile motion without the angle, use the vertical motion information to calculate time of flight, then use horizontal distance and time to find horizontal velocity, and combine it with initial vertical velocity using the Pythagorean theorem.

Step-by-step explanation:

To find the initial velocity in projectile motion without knowing the angle, one approach is to use the given vertical motion information. According to the provided discussion, if a projectile has an initial vertical velocity (Vy0) of 14.3 m/s and lands 20.0 m below its starting altitude, the time of flight is 3.96 seconds. This time is solely determined by the vertical motion, as the vertical acceleration is constant due to gravity.

Once the time of flight is known, you can calculate the horizontal component of the velocity (Vx) as follows: Vx = horizontal distance traveled / time of flight. It is important to note that Vx is constant throughout the motion if air resistance is negligible. The horizontal component can then be used in combination with the vertical component to calculate the total initial velocity (Vo). This can be done by using the Pythagorean theorem, Vo = √(Vx² + Vy0²).

Answer 2

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

`V= \sqrt{( (d)/(t))^(2)+( (gt)/(2))^(2) }`

Because g,t and d are known, V can be calculated.