Physics Study Problem Help Initial Speed and Horizontal Distance Given. Height?

Question

A racket ball is struck in such a way that it leaves the racket with a speed of 4.87m/s in the horizontal direction. When the ball hits the court, it is a horizontal distance of 1.95m from the racket. Find the height of the racket ball when it left the racket. I'm not sure what equation to use for this.

Thank you so much! I understand now how that fits together!

Answer 1

All answers to 3 sig figs.

t = x/V
... = 1.95/4.87
... = 0.400 sec

y = 1/2gt²
... = 1/2*9.81*0.400²
... = 0.785 m

Answer 2

You seem to have a mistake, as there is no equation for the y-component. The second equation: 0.300 m = 0.750 m + (11.7 m/s)(sin 18.5°)t − ½(9.80 m/s2)t2 does not contain yf. I believe the height of athlete's centre of mass is initially at 0.750m and finishes at 0.300m. This means : yf = 0.750 m + (11.7 m/s)(sin 18.5°)t − ½(9.80 m/s2)t^2 and yf =0.300m (b) A velocity v has horizontal and vertical components vcos(A) and vsin(A). From inspection of the equations (no calculation needed) you can see the velocity is 11.7m/s with A = 18.5°. It is wrong to say velocity is 11.6989 m/s as you cannot gve an answer to 6 significant figures when you have calculated it (which you needn't have) from data to only 3 significant figures. (c) You have to find the time in the air by solving the quadratic equation and then using this time to find xf. yf = 0.750 m + (11.7 m/s)(sin 18.5°)t − ½(9.80 m/s2)t^2 On landing yf =0.300m So we have to solve: 0.300m = 0.750 m + (11.7 m/s)(sin 18.5°)t − ½(9.80 m/s2)t^ Omitting units, for neatness: 0.750 + 11.7(sin(18.5)t − ½(9.80t^2 = 0.300 4.90t^2 -3.712t - 0.450 = 0 Solving the quadratic in the usual way gives: t = -0.1063s or 0.8639s Discard the negative solution as it is has no physical relevance xf = (11.7 m/s)(cos 18.5°) x 0.8639 = 9.59m