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!@
The vertical component of initial velocity is
Vyo = 45 m/s * sin 30 = 22.5 m/s
The time required to hit maximum height is t1 = yo/g and the time of flight is twice that.
The horizontal coordinate where the body hits the ground is
X = V cos 30 * (time of flight),
where V = 45 m/s
The maximum height is acquired when
Vyo = 0, at t = t1. The vertical height at that time is
V sin 30 * t1 - (g/2) t1^2
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!@
To determine the time of flight, distance traveled, and greatest height reached by the body, we can follow these steps using the given information:
1) Find the vertical component of the initial velocity (Vyo):
- Multiply the magnitude of the initial velocity (45 m/s) by the sine of the elevation angle (30 degrees):
Vyo = 45 m/s * sin(30) = 22.5 m/s
2) Calculate the time of flight (T):
- The time required to hit the maximum height is given by the formula: t1 = yo / g
(Where yo is the initial vertical displacement and g is the acceleration due to gravity)
- In this case, yo is not given explicitly, but since the body is projected upwards from the ground, yo can be assumed to be zero.
(Alternatively, if a value for yo is provided, it should be used in the calculation)
- Therefore, t1 = 0 / g = 0
- The time of flight is twice the time to reach maximum height:
T = 2 * t1 = 2 * 0 = 0 seconds
3) Determine the distance traveled (X) when the body strikes the ground:
- The horizontal coordinate where the body hits the ground is given by the formula: X = Vx * T
- Vx is the horizontal component of the initial velocity, which is the magnitude of the initial velocity (45 m/s) multiplied by the cosine of the elevation angle (30 degrees):
Vx = 45 m/s * cos(30) = 38.93 m/s
- X = 38.93 m/s * 0 = 0 meters.
(Note: This indicates that the body lands at the same horizontal position from where it was launched, assuming no external forces like air resistance or wind)
4) Find the greatest height reached (H):
- The greatest height is attained when the vertical component of the velocity becomes zero (Vyo = 0), at time t = t1.
- The vertical height at that time can be calculated using the formula: H = Vyo * t1 - (g / 2) * t1^2
- Substituting the values, H = 22.5 m/s * 0 - (9.8 m/s^2 / 2) * (0)^2 = 0 meters
(Note: This means that the maximum height reached is also zero, indicating that the body does not go higher than its initial elevation)
Therefore,
1) The time of flight is 0 seconds.
2) The distance from the point of projection to where the body strikes the ground is 0 meters.
3) The greatest height reached is 0 meters.
Please note that these calculations assume idealized conditions of uniform acceleration and neglect the effects of air resistance or any other external forces.