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.
To find the velocity of the stone, we can use the kinematic equation for projectile motion.
The equation for the vertical component of the projectile's motion is:
h = ut + (1/2)gt^2
where h is the height (in this case, 2m), u is the initial vertical velocity, t is the time taken, and g is the acceleration due to gravity, which is approximately -9.8 m/s².
Since the stone is thrown at an angle of 30° with the horizontal direction, let's find the vertical component of the initial velocity (u).
The vertical component can be calculated using:
u = v * sinθ
where v is the initial velocity of the projectile and θ is the angle of projection.
In this case, θ = 30°. So we have:
u = v * sin30°
Next, let's find the time taken (t) for the stone to reach the 2m height.
Using the equation:
h = ut + (1/2)gt^2
Substituting h = 2m and g = -9.8 m/s², we get:
2 = (v * sin30°) * t + (1/2)(-9.8)t^2
Now, let's solve this equation to find t.
Rearranging the equation, we get:
(1/2)(-9.8)t^2 + (v/2)t + 2 = 0
This is a quadratic equation with respect to t. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
where a = (1/2)(-9.8), b = (v/2), and c = 2.
By substituting the values in the quadratic formula, we can find two values of t. We can consider the positive value as the time taken for the stone to reach the height of 2m.
Once we have the value of t, we can find the initial velocity (v).
Let's solve the quadratic equation to find t, and then calculate v.
y = tanθ x - g/(2 (v cosθ)^2) x^2
so plug in your numbers and solve for v when
y(10) = 2