A body is projected such that the horizontal range is twice the maximum height of projection. Find the angle of projection; and the time of flight if the velocity of projection is 5m/s

vertical ... 2[5 sin(θ)] / g = t

... h = 5/2 sin(θ) * t/2

horizontal ... d = 5 cos(θ) t

2h = d
5/2 sin(θ) t = 5 cos(θ) t
dividing by cos(θ) t
and multiplying by 2/5
tan(θ) = 2

substitute back to find t

I don't really get you scott

Am I to substitute it this way 2[5(sin2)]/9.8.? Or will I make use of tan instead, since tan=2

Do tan-1 to find theta (how do you guys make a theta on this thing). Sub that back into the very first eq to find t

To find the angle of projection, as well as the time of flight, we can use the equations of motion for projectile motion.

Let's consider the given information:
- The horizontal range (R) is twice the maximum height (H)
- The velocity of projection (v) is 5 m/s

First, let's calculate the angle of projection (θ).

In projectile motion, the horizontal range (R) is given by the formula:
R = (v^2 * sin(2θ)) / g,
where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Since we know that R is twice the maximum height (H), we can write:
2H = (v^2 * sin(2θ)) / g.

Now, let's find the maximum height (H).
The maximum height (H) can be found using the formula:
H = (v^2 * (sin(θ))^2) / (2g).

Substituting this value of H into the equation 2H = (v^2 * sin(2θ)) / g, we get:
2 * (v^2 * (sin(θ))^2) / (2g) = (v^2 * sin(2θ)) / g.

Canceling out the common terms, we have:
(sin(θ))^2 = sin(2θ).

Using the double-angle formula for sine, we can rewrite the equation as:
sin(θ)^2 = 2 * sin(θ) * cos(θ).

Dividing both sides of the equation by sin(θ), we get:
sin(θ) = 2 * cos(θ).

Now, let's solve for θ:
Divide both sides of the equation by cos(θ):
tan(θ) = 2.

Using a calculator, find the inverse tangent (tan^-1) of 2:
θ ≈ 63.43 degrees.

So, the angle of projection is approximately 63.43 degrees.

Now, let's find the time of flight.

The time of flight (T) can be found using the formula:
T = (2*v*sin(θ)) / g.

Substituting the values, T = (2 * 5 * sin(63.43)) / 9.8, we can calculate the time of flight using a calculator:

T ≈ 1.05 seconds.

Therefore, the time of flight is approximately 1.05 seconds.

To summarize:
- The angle of projection is approximately 63.43 degrees.
- The time of flight is approximately 1.05 seconds.