when a stone is projected, it's horizontal range is 24m and greatest height 6m. find it's velocity of projection?
To find the velocity of projection of the stone, we can use the formulas for the range (horizontal distance) and maximum height of a projectile.
The formula for the range (R) is given by:
R = (v^2 * sin(2θ)) / g
where v is the initial velocity of projection, θ is the angle of projection, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The formula for the maximum height (H) is given by:
H = (v^2 * sin^2(θ)) / (2g)
We are given that the horizontal range is 24 m. This means R = 24.
We are also given that the greatest height is 6 m. This means H = 6.
Now, let's solve the equations.
From the equation for the maximum height (H), we can rearrange it to solve for v^2:
v^2 * sin^2(θ) = 2gH
v^2 = (2gH) / sin^2(θ)
From the equation for the range (R), we can rearrange it to solve for v^2:
R = (v^2 * sin(2θ)) / g
v^2 * sin(2θ) = Rg
v^2 = (Rg) / sin(2θ)
Now, from the equations for v^2, we can equate them:
(2gH) / sin^2(θ) = (Rg) / sin(2θ)
Cancel out the g terms on both sides:
2H / sin^2(θ) = R / sin(2θ)
Divide both sides by sin(2θ) and cross-multiply:
2H * sin(2θ) = R * sin^2(θ)
Apply the double-angle formula (sin(2θ) = 2sin(θ)cos(θ)):
4sin(θ)cos(θ) * H = R * sin^2(θ)
Divide both sides by sin(θ) and cancel out sin(θ):
4cos(θ) * H = R * sin(θ)
Divide both sides by cos(θ):
4H = R * tan(θ)
Now, we have an equation connecting the angle of projection (θ) to the range (R) and maximum height (H).
Next, we need to find the angle of projection (θ). To do this, we can use the inverse tangent function (tan^-1) on both sides of the equation:
θ = tan^-1((4H) / R)
Substituting the given values, θ = tan^-1((4*6) / 24) = tan^-1(1) = 45°
Finally, we can substitute the angle (θ) into the equation for velocity (v) to find the velocity of projection:
v^2 = (Rg) / sin(2θ)
v^2 = (24 * 9.8) / sin(90°)
v^2 = 235.2
v ≈ √235.2
v ≈ 15.33 m/s
Therefore, the velocity of projection of the stone is approximately 15.33 m/s.
To find the velocity of projection of a stone, we need to use the equations of motion. In this case, we have the horizontal range and the greatest height.
The horizontal range of a projectile is the total horizontal distance covered by the object during its motion. We can use the formula for horizontal range:
Range = (velocity × time of flight)
Given that the horizontal range is 24m, we have:
24 = (velocity × time of flight)
Next, we need to find the time of flight. The time of flight is the total time for which the projectile is in motion. We can use the formula for time of flight:
time of flight = 2 × (initial vertical velocity) / (acceleration due to gravity)
Given that the greatest height is 6m, we know that the vertical displacement at the highest point of the trajectory is zero. So, we can use the equation of motion for vertical displacement:
Vertical displacement = (initial vertical velocity × time of flight) - (0.5 × acceleration due to gravity × (time of flight)²)
Substituting the given values, we have:
6 = 0 - (0.5 × 9.8 × (time of flight)²)
Rearranging the equation, we get:
(time of flight)² = -(6 × 2) / (0.5 × 9.8)
Simplifying, we find:
(time of flight)² = -12 / 4.9
Taking the square root of both sides, we get:
time of flight = √(-12 / 4.9)
Since time cannot be negative, we ignore the negative sign and calculate the square root:
time of flight = √(12 / 4.9) ≈ 1.39 seconds
Now, substitute the value of the time of flight back into the equation for the horizontal range to solve for the velocity of projection:
24 = (velocity × 1.39)
Simplifying, we have:
velocity = 24 / 1.39
Calculating, we find:
velocity ≈ 17.27 m/s
Therefore, the velocity of projection of the stone is approximately 17.27 m/s.
v = Vi - g t
at top v = 0
so
Vi = 9.8 t
h = 0 + Vi t -4.9 t^2
6 = 0 + Vi t - 4.9 t^2
t is half the time in the air
6 = 9.8 t^2 - 4.9 t^2 = 4.9 t^2
so
t = sqrt (6/4.9) = sqrt (1.22) = 1.10 seconds upward
Vi = 9.8 (1.1) = 10.8 m/s initial velocity upward component
time in air = 2 t = 2.2 seconds
u (2.2 seconds) = 24
so u = 24/2.2 = 10.9 m/s horizontal speed
so
velocity = 10.9 i + 10.8 j
about 15 m/s at about 45 deg