Hi!
i need some help with this ASAP!
a body is projected upwards from a point on level ground with a velocity of 45 m/s at an elevation of 30 degrees. Working from the expression for uniformly accelerated motion in a straight line, determine
1) the time of flight
2) the distance from the point of projection to the point where the body strikes the ground
3) the greatest height reached
THANKS VERY MUCH!@
I will be happy to critiue your work or thinking. POst it, please.
ok
here goes..i want to know if the formula are kool
**** for the time of flight, i did this
= 2vsintheata / g
= 2(45sin30) / 9.81
= 45 / 9.81
= 4.59 s
**** for the distance, i did
= vcostheata * t (someone told me about this formula..is it correct???)
**** for the height, i did
h= ut + 1/2gt^2
u being the velocity which is 45m/s..rite...or 0??
that's my work thus far
Formulas? Formulas are no substitute for thinking.
<<a body is projected upwards from a point on level ground with a velocity of 45 m/s at an elevation of 30 degrees. Working from the expression for uniformly accelerated motion in a straight line, determine
1) the time of flight
2) the distance from the point of projection to the point where the body strikes the ground
3) the greatest height reached>>
Initial vertical velocity is 45m/s*sin30
at the top, the final vertical velcoty is -45m/s*sin30.
Vf= Vi - gt
solve for t, the time of flight.
Then for distance, distance=45*cosTheta*timeflight
heightmax= Vinitialvertical*timeup -1/2 g*timeup^2
where timeup= 1/2 time of flight.
One thinks in words, not formulas, and that is what I am trying to demonstrate. Most of your work is ok, as you will see when you go through my words. Rethink your height calcs.
To solve these problems, it is important to understand the concepts of projectile motion and uniformly accelerated motion.
1) Time of flight:
The time of flight is the total time it takes for the body to reach the ground after being projected upwards. To find the time of flight, you need to determine the time it takes for the body to reach its maximum height and then double that time.
First, let's determine the initial vertical velocity (Vi) of the body. Given that the velocity of 45 m/s is at an elevation of 30 degrees, we can find the vertical component of the velocity by multiplying it by the sine of the angle (sinθ).
Vi = 45 m/s * sin(30) = 22.5 m/s
Using the equation Vf = Vi - gt, where Vf is the final vertical velocity, g is the acceleration due to gravity (9.81 m/s^2), and t is the time, we can find the time it takes for the body to reach its maximum height.
Vf = -Vi (at the topmost point, the body momentarily stops before falling back down)
-22.5 m/s = 22.5 m/s - 9.81 m/s^2 * t
Simplifying the equation:
0 = 45 m/s - 9.81 m/s^2 * t
9.81 m/s^2 * t = 45 m/s
t = 45 m/s / 9.81 m/s^2
t ≈ 4.59 s
The time of flight is twice this value, so the time of flight is approximately 2 * 4.59 s = 9.18 s.
2) Distance from the point of projection to the point where the body strikes the ground:
To find the distance, we need to calculate the horizontal component of the velocity (Vx). This can be done by multiplying the initial velocity by the cosine of the angle (cosθ).
Vx = 45 m/s * cos(30) = 38.932 m/s
The distance (d) can be calculated using the formula distance = Vx * time of flight.
d = 38.932 m/s * 9.18 s
d ≈ 356.97 m
The distance from the point of projection to the point where the body strikes the ground is approximately 356.97 meters.
3) Greatest height reached:
To find the greatest height reached, we can use the equation h = Vi * t + (1/2) * g * t^2.
Substituting the values we have:
h = 22.5 m/s * (4.59 s / 2) + (1/2) * 9.81 m/s^2 * (4.59 s / 2)^2
h = 51.6825 m + 23.8866 m
h ≈ 75.57 m
Therefore, the greatest height reached is approximately 75.57 meters.
Remember, it's essential to understand the concepts and not just rely on formulas.