the horizontal range of a projectile is 4 root 3 times of its maximum height. calculate the angle of projection.
Let 'x' be the angle of projectile with the horizontal.
Then,taking g=10m/s^2:
Given, Hmax=[(usinx)^2]/2g;
R=4(root of 3)x(Hmax);
R=[(u^2)(sin2x)]/g;
Thus,
(u^2)sin2x 4(usinx)^2
-------- = --------------
g (root of 3)2g
Then,upon simplification:
sin2x 2(sinx)(sinx)
----- = -------------
1 (root of 3)
2sinxcosx 2
--------- = -----------
sinxsinx (root of 3)
cosx 1
---- = cotx = -----------
sinx (root of 3)
So,x=60 degrees.
It is not correct book answer
No satisfaction from the calculation shown
To calculate the angle of projection, we first need to understand the kinematic equations and the relationship between the horizontal range and maximum height of a projectile.
Let's define the given variables:
θ = angle of projection
R = horizontal range
H = maximum height
The horizontal range (R) can be calculated using the formula:
R = (v^2 * sin(2θ)) / g
where v is the initial velocity and g is the acceleration due to gravity.
The maximum height (H) can be calculated using the formula:
H = (v^2 * sin^2(θ)) / (2 * g)
Given that the horizontal range (R) is 4√3 times the maximum height (H), we can write the equation:
R = 4√3 * H
Now, let's solve for the angle of projection (θ).
Substituting the expressions for R and H, we have:
(4√3 * H) = (v^2 * sin(2θ)) / g
Rearranging the equation, we get:
(4√3 * H * g) = v^2 * sin(2θ)
To simplify the equation further, we can use the double-angle identity for sine:
sin(2θ) = 2 * sin(θ) * cos(θ)
Replacing sin(2θ) with its equivalent, the equation becomes:
(4√3 * H * g) = v^2 * (2 * sin(θ) * cos(θ))
Simplifying the equation, we get:
2v^2 * sin(θ) * cos(θ) = 4√3 * H * g
Dividing both sides by 2v^2 * g, we have:
sin(θ) * cos(θ) = (2√3 * H) / v^2
Using the identity sin(2θ) = 2 * sin(θ) * cos(θ), we can substitute sin(2θ) with its equivalent:
sin(2θ) = (2√3 * H) / v^2
Now, we can find θ by taking the inverse sine (sin^-1) of both sides:
2θ = sin^-1((2√3 * H) / v^2)
Dividing both sides by 2, we obtain:
θ = (1/2) * sin^-1((2√3 * H) / v^2)
Therefore, to calculate the angle of projection, you need to know the values of the initial velocity (v) and the maximum height (H). Substitute these values into the equation above and solve for θ.