An object is projected with velocity of 100m/s from the ground level at an angle tita at the vertical,if the total thme is 10s.calculate tita(g=10m/s)
suppose a string is extended by 0.2 by the application of the force of 10m. calculate the elastic constant
lesson4
To calculate the angle θ at which the object is projected, given the initial velocity, total time, and gravitational acceleration, you can use the equations of motion.
We can start by breaking down the initial velocity into its horizontal and vertical components. The horizontal component (Vx) remains constant throughout the motion, while the vertical component (Vy) changes due to the effect of gravity.
Given:
Initial velocity (Vo) = 100 m/s
Total time (t) = 10 s
Gravitational acceleration (g) = 10 m/s²
1. Calculate the time of flight:
The total time (t) is equal to the sum of the times taken to reach the maximum height (t₁) and return to the ground (t₂).
t = t₁ + t₂
Since the object reaches its maximum height halfway through its motion, t₁ = t / 2 and t₂ = t / 2.
So, t = t / 2 + t / 2
=> t = 2 * t / 2
=> t = 2 * t₁
Therefore, t₁ = t / 2 = 10 s / 2 = 5 s.
2. Calculate the maximum height (h):
The maximum height can be calculated using the formula:
h = Vyo * t₁ - (1/2) * g * t₁²
Since the initial vertical velocity (Vyo) is given by Vyo = Vo * sin(θ), we can substitute the value for Vyo and rearrange the equation to solve for sin(θ).
h = (Vo * sin(θ)) * (t / 2) - (1/2) * g * (t / 2)²
After substituting the given values, we have:
h = (100 * sin(θ)) * (10 / 2) - (1/2) * 10 * (10 / 2)²
Simplifying further:
h = 50 * sin(θ) - 25
3. Solve for sin(θ):
To solve for sin(θ), we need to rearrange the equation and isolate sin(θ) on one side:
h + 25 = 50 * sin(θ)
Divide both sides by 50:
(sin(θ)) = (h + 25) / 50
Take the inverse sine (sin⁻¹) of both sides to isolate θ:
θ = sin⁻¹((h + 25) / 50)
Now, plug in the value of h to calculate the angle θ.