I need help with part a, b, c, d, and e. Thank you. 1. A ball is launched from the ground with an initial velocity of 6.4 m/s at an angle of 60º above thehorizontal as shown below and lands in a bucket located 2.1 meters away. Determine;a. The horizontal distance traveled by the ball at t = 0.35 sec.b. The height of the bucket.The magnitude of Vy when the ball reaches maximum height.d. The magnitude of the ball's velocity just before it lands in the bucket.Is the ball rising or falling at t = 0.35 seconds? Show the calculation in support of youranswer.Vo600

As per guidelines , only first 3 subparts are solved here. Given, Ball is launched with initial velocity 6.4m/s at angle of 60° above the horizontal and lands in a bucket located 2.1 m away. u = 6.4m/sθ = 60°Horizontal acceleration is 0 m/s2 and vertical acceleration is acceleration due to gravity = 9.8 m/s2 a. Horizontal distance = Horizontal velocity×time , as horizontal acceleration is 0. Horizontal velocity = ucosθ Horizontal distance at time t = 0.35 s , = (6.4m/s)(cos60°)(0.35s) =1.12 m b. The height of bucket is given as y = xtanθ+gx22u2(cos2θ) , where y is height of bucket and x = 2.1 mso, height of the bucket =2.1m(tan60°)+9.8m/s2×(2.1m)22×(6.4m/s)2(cos260°) =5.75m c. At maximum height , vertical velocity becomes 0 m/s as at maximum height , the ball has only horizontal component of velocity. So, vy = 0 m/s