A projectile was launched 64° above the horizontal, attaining a height of 10 m. What is the projectile's initial speed?
To find the initial speed of the projectile, we can use the laws of projectile motion. The key equation for projectile motion is the horizontal range equation:
R = (v^2 * sin(2θ)) / g
where:
- R is the range (the horizontal distance traveled by the projectile),
- v is the initial speed of the projectile,
- θ is the launch angle (64° in this case),
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this problem, we need to find v. We are given the launch angle (θ = 64°) and we know the projectile reaches a height of 10 m. However, we do not have the horizontal range.
To solve this, we can use the fact that the time taken to reach maximum height and the time taken to fall back to the ground are equal in symmetrical projectile motion (launch and landing at the same height). Therefore, the time taken to reach maximum height is the same as the total flight time, which we can calculate using the vertical motion equations.
The equation for the projectile's maximum height, H, is:
H = (v^2 * sin^2(θ)) / (2g)
In this problem, we are given H = 10 m and θ = 64°. Plugging in these values, we can solve for v.
10 = (v^2 * sin^2(64°)) / (2 * 9.8)
Next, we can rearrange this equation to solve for v^2:
v^2 = (10 * 2 * 9.8) / sin^2(64°)
Finally, we can take the square root of both sides to find the initial speed of the projectile:
v = √((10 * 2 * 9.8) / sin^2(64°))
Evaluating this expression will give us the answer, which is the initial speed of the projectile.
(Average vertical velocity) * (time of flight to max height) = Maximum height H
(Vo*sinA/2)*(VosinA/g) = H
Vo^2*sin^2A/(2g) = H
Solve for Vo