an arrow is shot into air. its range is 200m and its time of flight is 5s . if g=10m/s2 ,then horizontl compoent of velocity and the mximum height will be
To find the horizontal component of velocity and the maximum height of the arrow, we can use the equations of motion for projectile motion.
The horizontal component of velocity remains constant throughout the motion, so we can use the formula:
Range = Horizontal component of velocity × Time of flight
Given that the range of the arrow is 200m and the time of flight is 5s, we can rearrange the equation to solve for the horizontal component of velocity:
Horizontal component of velocity = Range / Time of flight
Substituting the given values into the equation:
Horizontal component of velocity = 200m / 5s = 40m/s
Therefore, the horizontal component of velocity is 40m/s.
To find the maximum height, we can use the formula for vertical displacement in projectile motion:
Vertical displacement = (Initial vertical component of velocity × Time of flight) - (0.5 × acceleration due to gravity × Time of flight^2)
Since the arrow was shot vertically upwards, the initial vertical component of velocity is equal to the final vertical component of velocity, and it is 0 m/s when the arrow reaches its maximum height.
So, the equation simplifies to:
Vertical displacement = - (0.5 × acceleration due to gravity × Time of flight^2)
Given that the acceleration due to gravity (g) is 10 m/s^2 and the time of flight is 5s, we can substitute these values into the equation:
Vertical displacement = - (0.5 × 10 m/s^2 × (5s)^2)
Vertical displacement = - (0.5 × 10 m/s^2 × 25s^2)
Vertical displacement = - (0.5 × 10 m/s^2 × 625)
Vertical displacement = - (5 m/s^2 × 625)
Vertical displacement = - 3125 m
However, since the maximum height is measured from the ground level, the vertical displacement should be positive. Therefore, we take the absolute value of the calculated vertical displacement:
Maximum height = | - 3125 m | = 3125 m
Therefore, the maximum height of the arrow is 3125m.
To find the horizontal component of velocity and the maximum height, we can use the equations of projectile motion. Let's denote the horizontal component of velocity as Vx and the vertical component of velocity as Vy.
1. Finding the horizontal component of velocity (Vx):
We know that the range (R) of the projectile is given by the formula:
R = Vx * time of flight
Substituting the given values into the formula:
200m = Vx * 5s
Divide both sides by 5s:
Vx = 200m / 5s
Vx = 40 m/s
Therefore, the horizontal component of velocity (Vx) is 40 m/s.
2. Finding the maximum height:
The time of flight (T) can be used to find the time taken to reach the maximum height (T/2).
T/2 = 5s / 2
T/2 = 2.5s
Using the formula for the time of flight in terms of vertical velocity (Vy) and gravitational acceleration (g):
T = 2 * Vy / g
Substituting the given value of T/2:
2.5s = 2 * Vy / 10m/s^2
Multiply both sides by 10m/s^2:
25m/s = 2 * Vy
Divide both sides by 2:
12.5m/s = Vy
Therefore, the vertical component of velocity (Vy) is 12.5 m/s.
Now, to find the maximum height (H), we can use the formula:
H = Vy^2 / (2 * g)
Substituting the given values:
H = (12.5m/s)^2 / (2 * 10m/s^2)
H = 156.25m / 20m/s^2
H = 7.8125m
Therefore, the maximum height reached by the arrow is 7.8125 meters.
1. R = Xo * T = 200 m.
Xo*5 = 200
Xo = 40 m/s. = Hor. component of initial
velocity.
2. Tr = T/2 = 5 / 2 = 2.5 s.
Y = Yo + gt.
Yo = Y - gt = 0 - (-10)*2.5 = 25 m/s. =
Ver. component of initial velocity.
hmax = (Y^2-Yo^2)/2g.
hmax = (0-(25)^2) / -19.6 = 31.9 m.