Question

Pleasd help and explain if possible. Thanks!

Menu Principal Contenidos Grades Contenido de Curso > ... > Cap4. Movimiento en 2D y 3D - Timer Notas Evaluate Feedback 2 Imprimir Serving at a speed of 155 km/h, a tennis player hits the ball at a height of 2.43 m and an angle below the horizontal. The player serves at a distance of 12.2 m from the net, which is 0.91 m high. What is the angle such that the ball just crosses the net? 5.26 deg ¡Correcto! Su recibo es 160-309 Intentos Anteriores The outline of the service box is 6.40 m from the net. How far does the ball land from the outline of the service box? (If the ball lands outside the outline, enter your answer as a positive number. If inside, enter as a negative number.) -17.49 m Enviar Respuesta Incorrecto. Tries 11/40 Intentos Anteriores e Enviar Mensaje de Discusión Enviar Reacción

Answer 1

A u= 155 km/h = 43 m/s hi =2-43 m, h = 0.91m xi = 12. 2m 4w hn) I ti k e toutlina let time to reach net be t o then not vertical displacement to not in Bi -ħn = usino t + tgt? and xi = ucaso xt + 12.2 = 43 cane xt t = 0.284 case Purting this value of t in O. bị - An = (u Ane 6 1284 + 4.9 xe 294) Tase caró 32.43.0.91= 12-2 tano + 0.395 (It tan o) → 0.395 temo + 12.2 temo - 1+125=0 > tamo z 0-991 50 = 50199! trand=can t-0-284 Titremos como

The outline of service box is 6.4m from net when ball lands on around then let total time of flight be to hi = usino x T +491 ☺ and Gita) = ucaso x T (Total horizontal from eq o distance ) 2:43 = 43 sin 5.2°x T +1x9.81*T => 2:43 = 3.897 T +4-972 =) T = 0.411 sec. and (xitxa) = (43 cas 5. 2X0.411 - 12.2 + x = 17.6 => x = 5.4 m. This ball entors lands imide of the outline of the service box. by a distance. (604-5.4) = 1 mi Thus the answer is -Im lame wer