In a projectile motion, the horizontal range and the maximum height attained by the projectile are equal .
A) what is the launch angle?
B) If everything else slays the same , how should the launch angle, ¦È0, of a projectile be changed for the range of the projectile to be halve?
be halved??
recall that
range = v^2/g sin(2θ)
height = v^2/g sin^2(θ)
so, solve for θ
an excellent discussion of trajectories for all your problems can be found at
https://en.wikipedia.org/wiki/Trajectory
thanks a lot sir@Steve
To find the launch angle in a projectile motion, we need to analyze the components of the motion separately: the horizontal component and the vertical component.
A) To determine the launch angle when the horizontal range is equal to the maximum height, we can use the fact that the time taken to reach the maximum height is half the total time of flight.
Let's assume that the initial velocity of the projectile is v0 and the launch angle is ¦È0.
The horizontal component of the projectile's velocity remains constant throughout the motion, so we can find the time of flight (t) using the formula:
t = 2 * (horizontal range) / (horizontal velocity)
Since the horizontal range is equal to the maximum height, we can write:
t = 2 * (maximum height) / (v0 * sin(¦È0))
On the other hand, the time taken to reach the maximum height only involves the vertical component of the projectile's velocity. Using the formula:
t = (vertical velocity) / (acceleration due to gravity)
Since the vertical velocity at the maximum height is zero and the acceleration due to gravity is constant, we can write:
t = v0 * sin(¦È0) / g
Equating the two expressions for t, we get:
2 * (maximum height) / (v0 * sin(¦È0)) = v0 * sin(¦È0) / g
Canceling out common terms and rearranging the equation, we can solve for the launch angle, ¦È0:
2 * maximum height * g = v0² * sin(2¦È0)
sin(2¦È0) = 2 * maximum height * g / v0²
Using the inverse sine function, we can find the value of 2¦È0:
2¦È0 = arcsin(2 * maximum height * g / v0²)
Therefore, the launch angle, ¦È0, can be determined as:
¦È0 = 0.5 * arcsin(2 * maximum height * g / v0²)
B) If we want to halve the range of the projectile while keeping everything else the same, we can use the same formula but with the new range value:
0.5 * (maximum height) * g = v0² * sin(2¦È1)
where ¦È1 is the new launch angle.
Now, we can solve for ¦È1 using the same steps as before:
2¦È1 = arcsin(0.5 * maximum height * g / v0²)
¦È1 = 0.5 * arcsin(0.5 * maximum height * g / v0²)
Therefore, the launch angle, ¦È1, should be equal to half the launch angle, ¦È0, for the range of the projectile to be halved.