A soccer player kicked a ball at an angle of 37 to the horizontal if the range covered by the ball is 240.3m calculate
What is the initial velocity of projection?
The time of flight?
The maximum leight attained
recall that the range is given by
R = v^2/g sin2θ
So,
v^2 = 9.8*240.3/sin74° = 2449.8
v = 49.5 m/s
Now you can use that to find the other two values.
To find the initial velocity of projection, time of flight, and maximum height attained by the ball, we can use the equations of projectile motion.
1. Initial velocity of projection (u):
The horizontal range covered by the ball is given by the equation:
Range = (u^2 * sin(2θ)) / g
Where,
- Range is the horizontal distance covered by the ball (240.3m).
- u is the initial velocity of projection.
- θ is the angle of projection (37 degrees).
- g is the acceleration due to gravity (9.8 m/s^2).
Rearranging the equation to solve for u:
u^2 = (Range * g) / sin(2θ)
Now we can substitute the given values and calculate the initial velocity:
u^2 = (240.3 * 9.8) / sin(2 * 37)
u^2 ≈ 2464.36
Taking the square root of both sides:
u ≈ √2464.36
u ≈ 49.64 m/s
Therefore, the initial velocity of projection is approximately 49.64 m/s.
2. Time of flight (T):
The time of flight is the total time taken by the ball to reach the ground. It can be calculated using the equation:
T = (2 * u * sin(θ)) / g
Substituting the given values:
T = (2 * 49.64 * sin(37)) / 9.8
T ≈ 5.14 seconds
Therefore, the time of flight is approximately 5.14 seconds.
3. Maximum height attained (H):
The maximum height attained by the ball can be calculated using the equation:
H = (u^2 * sin^2(θ)) / (2 * g)
Substituting the given values:
H = (49.64^2 * sin^2(37)) / (2 * 9.8)
H ≈ 20.16 meters
Therefore, the maximum height attained by the ball is approximately 20.16 meters.