Physics, 24.09.2019 12:50 kelsea86

How to find initial velocity in projectile motion without angle?

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Answer from: starreee

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
$V= \sqrt{( \frac{d}{t})^{2}+( \frac{gt}{2})^{2} }$

Because g,t and d are known, V can be calculated.
How to find initial velocity in projectile motion without angle?

How to find initial velocity in projectile motion without angle?

Answer from: Quest

b mirrors submit a iisner? e. for which optical devices does d sometimes have a positive value? o a a and b o b. a and c o c. b and d 0 d. b and cb

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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