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.