Find the maximum height and the time of flight of a body fired vertically up from the ground given that it loses 60 percent of its initial speed in rising 4.2m
At 4.2 [m], it loses 60% of its initial speed, u. This means that at 4.2 [m], the speed, v = u - 0.6u.
So, by using v^2 = u^2 - 2gs, assume g = 9.81 [m/s^2], s = 4.2 [m]
(u - 0.6u)^2 = u^2 - 2gs
u = 9.9045 [m/s], this is the initial speed.
Then, at the maximum height, v = 0. So from v = u - gt, at the maximum height,
0 = u - gt, substitute initial speed,u, and g = 9.81 [m/s^2]
the time at the maximum height, t = 1.0096 [s]. For the time of flight of a body fired vertically up from the ground is the double value of the time at the maximum height.
So, the time of flight of a body fired vertically up from the ground is 1.0096 x 2 = 2.0193 [s]
Then, to find the maximum height, by using s = ut - 0.5gt^2, where t = 1.0096 [s],
s = ut - 0.5gt^2 = 9.9045(1.0096) - 0.5(9.81)(1.0096)^2
= 9.9996 - 4.9996 = 5 [m]
Aaz=Emanuel?
v = Vi - g t
h = .5 g t^2
at 4.2 meters, t=time at 4.2m
4.2 = .5 g t^2
t = sqrt (8.4/g)
v = Vi - sqrt (8.4 g) = 0.6 Vi
so
0.4 Vi = sqrt(8.4 g)
Vi = sqrt(8.4 g) / 0.4
use 9.81 or whatever you use for g and you have the initial speed up
0 = Vi - g t finds t at top
double that for total time in air
h = .5 g t^2 using t at top
Well, if the body is fired vertically up, it's quite a unique way to make your entrance! And losing 60 percent of its speed in rising just 4.2 meters? That's quite the show-off move. But hey, I'm here to help!
Let's figure out the maximum height and the time of flight for this airborne acrobat. First, we need to find the initial speed of the body. Since it loses 60 percent of its speed in rising 4.2 meters, we can assume that the body reaches its maximum height when it has lost 60 percent of its initial speed.
So, let's call the initial speed of the body "V." If it loses 60 percent of its speed, it means it retains 40 percent. Therefore, the speed at the maximum height is 40 percent of V.
Now, in order to find the maximum height, we need to use the concept of conservation of energy. At the maximum height, the body's kinetic energy (due to speed) is zero, and all its initial energy is in the form of potential energy (due to height).
So, let's use the equation:
Potential energy at the maximum height = Initial kinetic energy - Energy lost in rising 4.2 meters
Taking the formulas for kinetic and potential energy into account, we can write:
mgh = (1/2)mv^2 - (0.6)mv^2
Here, m is the mass of the body, g is the acceleration due to gravity, and h is the maximum height.
Simplifying the equation, we get:
gh = (1/2)(1 - 0.6)v^2
Since g, h, and v are all positive values, we can drop them from both sides of the equation to get:
h = (1/2)(1 - 0.6)v^2
Now, to find the time of flight, we can use the formula:
Time of flight = 2*(Initial speed)/(Acceleration due to gravity)
Substituting the value of the initial speed (40 percent of V) and the acceleration due to gravity, we get:
Time of flight = 2*(40 percent of V)/g
So, using these equations, you can calculate the maximum height and the time of flight for this daring body. Enjoy the performance!
To find the maximum height and the time of flight of a body fired vertically up from the ground, we need to use the equations of motion. Let's break down the problem step by step:
Step 1: Define the known values:
- Initial speed (u) is the speed at which the body is fired vertically up.
- The loss of speed is 60 percent of the initial speed, equals 0.6u.
- The height lost (h) during the deceleration is 4.2m.
Step 2: Determine the equations to use:
- The equation for velocity is v = u - gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is time.
- The equation for height is h = ut - (1/2)gt^2, where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is time.
Step 3: Find the time of deceleration:
- During deceleration, the final velocity (v) is zero. Substituting the given values into the velocity equation, we get 0 = u - gt.
- Rearranging the equation, we can find t: t = u / g.
Step 4: Find the deceleration distance:
- Substitute the found value of time (t) into the height equation: h = ut - (1/2)gt^2.
- Rearrange the equation to express the deceleration distance, h: h = u(u/g) - (1/2)g(u/g)^2.
Step 5: Solve for the deceleration distance:
- Substitute the value of the deceleration distance, h = 4.2m, into the equation.
- Simplify the equation: 4.2 = u(u/g) - (1/2)g(u/g)^2.
Step 6: Solve for initial speed (u):
- Rearrange the equation: 4.2 = u^2/g - (1/2)u^2/g.
- Combine terms: 4.2 = (1/2)u^2/g.
- Multiply both sides by 2g: 8.4g = u^2.
- Take the square root of both sides: u = √(8.4g).
Step 7: Find the maximum height:
- Substitute the found value of initial speed (u) into the height equation: h = ut - (1/2)gt^2, where t = 2(u/g) since the body goes up and comes back down.
- Simplify the equation: h = 2(u/g)(u/g) - (1/2)g(2(u/g))^2.
- Simplify further: h = 2u^2/g - 2g(u/g)^2.
- Substitute the value of u from step 6: h = 2(8.4g)/g - 2g(8.4g/g^2).
- Simplify the equation: h = 16.8 - 16.8.
- The maximum height is 0m.
Step 8: Find the time of flight:
- Given that the maximum height is 0m, it means the body goes up and immediately comes back down. The time of flight is twice the time of deceleration, t = 2(u/g).
- Substitute the value of u from step 6: t = 2(√(8.4g)/g).
- Simplify the equation: t = 2(2.9)√g.
- The time of flight is 5.8√g.
Therefore, the maximum height is 0m, and the time of flight is 5.8√g, where g is the acceleration due to gravity.