A soccer player kicked a ball at an angle of 37 degrees to the horizontal.if the range covered by the ball is 240.3m; calculate
(a) The initial velocity of projection ?
(b) The time of flight
(c) The maximum height attained
To solve this problem, we can break it down into three parts: finding the initial velocity (v₀), finding the time of flight (t), and finding the maximum height (h).
(a) To find the initial velocity (v₀), we need to use the horizontal range formula:
Range = (v₀² * sin(2θ)) / g
Given: Range = 240.3m and θ = 37 degrees
Rearranging the formula, we get:
v₀² = (Range * g) / sin(2θ)
Substituting the values, we have:
v₀² = (240.3 * 9.8) / sin(2 * 37)
Now, calculate sin(2 * 37) and solve for v₀:
sin(74) = sin(2 * 37) = 0.96592582628
v₀² = (240.3 * 9.8) / 0.96592582628
v₀² = 2445.86
Taking the square root of both sides, we find the initial velocity:
v₀ = √(2445.86)
v₀ ≈ 49.455 m/s
Therefore, the initial velocity of projection is approximately 49.455 m/s.
(b) To find the time of flight (t), we can use the vertical motion formula:
t = (2 * v₀ * sin(θ)) / g
Given: v₀ = 49.455 m/s and θ = 37 degrees
Substituting the values, we have:
t = (2 * 49.455 * sin(37)) / 9.8
t = 3.03 seconds (rounded to two decimal places)
Therefore, the time of flight is approximately 3.03 seconds.
(c) To find the maximum height (h), we can use the vertical motion formula:
h = (v₀² * sin²(θ)) / (2 * g)
Given: v₀ = 49.455 m/s and θ = 37 degrees
Substituting the values, we have:
h = (49.455² * sin²(37)) / (2 * 9.8)
h ≈ 29.7 meters
Therefore, the maximum height attained is approximately 29.7 meters.