Answer :
a) Initial velocity of the ball u = 30 m/sAngle of projection of the ball, θ = 41°In the vertical direction, s = ? and u = 30 m/s, θ = 41°, g = 9.8 m/s²Vertical motion of the ball is given by,s = ut sin θ - ½ gt²Maximum height is reached when vertical velocity is zero.
So at highest point v = 0Therefore, final velocity v = 0s = ut sin θ - ½ gt² 0 = 30 × sin 41° - ½ × 9.8 × t² ⇒ t = 3.40 sNow using the formula, s = ut sin θ - ½ gt² ⇒ s = 30 × sin 41° × 3.40 - ½ × 9.8 × 3.40²Hence, the maximum height that the ball reaches if the initial speed is 30 m/s is 49.7 m.b) The distance covered by the ball will be 98.1 m.According to the given problem, the ball will land at the same height as it started.
So, net displacement of the ball in the vertical direction will be zero.Hence, we can use the horizontal range formula to calculate initial velocity of the ball.R = u² sin 2θ/gHere, range R = 98.1 m, θ = 41° and g = 9.8 m/s²∴ 98.1 = u² sin 82°/9.8 u = 98.1 × 9.8/sin 82° ≈ 128 m/sHence, the initial speed of the ball should be approximately 128 m/s.c) To kick the ball with the smallest initial speed to the other end of the field, we must choose the angle such that the horizontal range is maximum. Here, the given field has a fixed length.
So, we must select the angle in such a way that the range is maximum but less than or equal to the length of the field.The range is maximum when θ = 45°.R = u² sin 2θ/gHere, θ = 45° and g = 9.8 m/s²∴ R = u² sin 90°/9.8 = u²/9.8Since R = 98.1 m, u²/9.8 ≤ 98.1 u² ≤ 961.38 u ≤ 31.01 m/sHence, the angle of projection should be 45° to kick the ball to the other end of the field with the smallest initial speed.
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