# 1-kg ball at A is suspended by an inextensible cord and given an initial horizontal velocity of 5...

## Question:

1-kg ball at A is suspended by an inextensible cord and given an initial horizontal velocity of 5 m/s. If / = 0.62 m and xB = 0, determine yB so that the ball will enter the basket. (Round the final answer to three decimal places.)

## Finding Height Of The Ball:

To find the height of the ball we need to use energy conservation to get the velocity of a coordinate and then apply Newtons law of motion for a respective coordinate where the distance is to be found.

v= 5 m/s.

L= 0.62 m.

We can write for y- direction

{eq}mgsin\theta=\frac{mv_2^2}[L}\\ v_2^2=\sqrt(gLsin\theta)\\ v_2^2=\sqrt(9.8\times 0.62 sin\theta) {/eq}

Conservation of energy

{eq}\frac{1}{2}mv_1^2+mgh_1=\frac{1}{2}mv_2^2+mgh_2\\ \frac{1}{2}(1)(5)^2+9.8(-0.62)=\frac{1}{2}(1)\sqrt(9.8\times 0.62\times sin\theta)^2+9.8(0.62)sin\theta\\ \theta=sin^{-1}(0.704)=44.81^0 {/eq}

Finding time.

{eq}l cos\theta=v_2sin\theta\times t\\ t=\frac{L cos\theta}{v_2 sin\theta}\\ t=\frac{0.62}{2.06}cot (44.81)\\ t=0.301 {/eq}

Using Newtons equation.

{eq}y_B=x+v_2-\frac{1}{2}gt^2\\ =0.62 sin 44.8+2.06 cos 44.8(0.301)-\frac{1}{2} 9.8(0.301)^2\\ =0.4329 \ m {/eq} 