A 1.3 kg ball at the end of a 1.2 m string swings in a vertical plane. At its lowest point the ball is moving with a speed of 14 m/s.?

a) use v² = u² + 2gh

v² = (14m/s)² + 2 * -9..8m/s² * 2*1.2m = 149 m²/s² → v = 12 m/s

b) T = m*(v²/r ± g)

at bottom, T = 1.3kg * (14²/1.2 + 9.8)m/s² = 225 N

at top. T = 1.3kg * (12²/1.2 - 9.8)m/s² = 149 N

Hope this helps!

Let the velocity of the ball is v m/s at the top, By the law of energy conservation,

=>KE(bottom) = KE(top) + PE(top)

=>1/2m(14)^2 = 1/2mv^2 + mg x (2 x 1.2)

=>196 = v^2 + 47

=>v = √149

=>v = 12.21 m/s

(a) Let the Tension in the string is T Newton at the top, by balancing the forces

=>mv^2/r - mg = T

=>T = [1.3 x (12.21)^2]/1.2 - 1.3 x 9.8

=>T(top) = 148.68 N

(b) Let the Tension in the string is T Newton at the bottom, by balancing the forces

=>mv^2/r + mg = T

=>T = [1.3 x (14)^2]/1.2 + 1.3 x 9.8

=>T(top) = 225.07 N