The highest barrier that a projectile can clear is 14.5 m, when the projectile is launched at an angle of 31.0 ° above the horizontal. What is the projectile's launch speed?
Only the vertical component of velocity coverts to gravitational potential energy at maximum height H.
(Vo sin31)^2 = 2 g H = 284.2 m^2/s^2
Vo sin31 = 16.86 m/s
Vo = 32.7 m/s
To find the projectile's launch speed, we can use the equation for the maximum height of a projectile. The maximum height is reached when the vertical component of the velocity becomes zero.
The equation for the maximum height is given by:
h = (v^2 * sin^2(θ)) / (2 * g)
where:
h = maximum height
v = launch velocity
θ = launch angle
g = acceleration due to gravity (approximated as 9.8 m/s^2)
In this case, the maximum height is given as 14.5 m and the launch angle is 31.0°.
Substituting these values into the equation, we have:
14.5 = (v^2 * sin^2(31.0°)) / (2 * 9.8)
To solve for v, we need to rearrange the equation:
v^2 = (14.5 * 2 * 9.8) / sin^2(31.0°)
v^2 = 285.8 / (sin^2(31.0°))
Taking the square root of both sides, we get:
v = √(285.8 / (sin^2(31.0°)))
Using a calculator, we can solve for v:
v ≈ √(285.8 / (0.1543^2)) ≈ 36.3 m/s
Therefore, the projectile's launch speed is approximately 36.3 m/s.
To find the projectile's launch speed, we can use the concepts of projectile motion and the equations of motion.
The first step is to break the motion of the projectile into its horizontal and vertical components.
1. Horizontal Component: The horizontal velocity of the projectile remains constant throughout its motion. We can find the horizontal component of the initial velocity using the formula:
vx = v * cos(θ), where vx is the horizontal component of velocity, v is the initial velocity, and θ is the launch angle.
In this case, θ = 31.0 °.
2. Vertical Component: The vertical motion of the projectile is influenced by gravity. The vertical displacement (height) of the projectile can be calculated using the formula:
h = vy * t + (1/2) * g * t^2, where h is the maximum height, vy is the vertical component of velocity, t is the time of flight, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
At the maximum height, vy = 0, so the equation becomes:
h = (1/2) * g * t^2.
3. Time of flight: The time it takes for the projectile to reach its maximum height and return to the same vertical position can be calculated using the formula:
t = 2 * vy / g.
At the maximum height, vy = 0, so the equation becomes:
t = 2 * (v * sin(θ)) / g.
To find the launch speed, we need to solve for v.
First, solve for t using the equation t = 2 * (v * sin(θ)) / g:
t = 2 * (v * sin(31.0 °)) / 9.8
Then, substitute t in the equation h = (1/2) * g * t^2:
14.5 = (1/2) * 9.8 * (2 * (v * sin(31.0 °)) / 9.8)^2
Simplify the equation:
14.5 = (v^2 * sin^2(31.0 °)) / 9.8
Multiply both sides by 9.8 to eliminate the denominator:
14.5 * 9.8 = v^2 * sin^2(31.0 °)
Take the square root of both sides to solve for v:
v = sqrt((14.5 * 9.8) / sin^2(31.0 °))
Using a calculator, we can find the value of v to be approximately 19.5 m/s.
Therefore, the projectile's launch speed is approximately 19.5 m/s.