a stone is thrown at an angle of 30* with the horizontal direction which strikes the tip of the building 2 m height ,if the distance of the building from the point of observation is 10 metre .find the velocity of the projectile of the stone.
ans...
To find the velocity of the projectile, we can use the equation of motion for projectile motion in the vertical direction:
h = (u^2 * sin^2θ) / (2g)
Where:
h = height of the building (2 m)
u = initial velocity of the stone
θ = angle of projection (30 degrees)
g = acceleration due to gravity (9.8 m/s^2)
Since the stone strikes the tip of the building, the vertical displacement is equal to the height of the building.
Substituting the given values into the equation, we have:
2 = (u^2 * sin^2(30)) / (2 * 9.8)
Simplifying the equation, we get:
u^2 * sin^2(30) = 2 * 9.8 * 2
u^2 * (1/4) = 39.2
u^2 = 157.6
Taking the square root of both sides, we get:
u ≈ 12.56 m/s
Therefore, the velocity of the projectile stone is approximately 12.56 m/s.