Question

please answer both questions.

My Notes Ask Your Teacher asketball star covers 2.55 m horizontally in a jump to dunk the ball. His motion through space can be modeled precisely as at of a particle at his center of mass. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.80 m above the floor and is at elevation 0.950 m when he touches down again. (a) Determine his time of night (his "hang time"). (b) Determine his horizontal velocity at the instant of takeoff. m/s (c) Determine his vertical velocity at the instant of takeoff. m/s (d) Determine his takeoff angle. above the horizontal (e) For comparison, determine the hang time of a whitetail deer making a jump with center-of-mass elevations Y = 1.20 m, max = 2.40 m, and y, = 0.670 m. Need Help? Raadt -/10 points SERPSE10 4.3.P.009. My Notes Ask Your Teacher V The speed of a projectile when it reaches its maximum height is 0.43 times its speed when it is at half its maximum height. What is the initial projection angle of the projectile?

Answer 1

Ano Ox= .55 m yi - 1.02 m 'm a = 1. 8 . 4 > 0.95 m. At max point iti velocity is zero and only gravity acts vertical direction ti: time to go from yi to Yomax > Vy = Uy - gti owy-gt, => Wy=gt. - 0 Now using a h=ut + lat => ymax = y; = Uy ti + 149)t? 0.78 - gti-ati? +0.75 = qui* = 9,8 ti? → 0.78 =gt? 2 => t = 0.395 — . Fi d₂: time to go from Ymax to yf using ☺ again = Y; - Ymax = 0 t2 - 9+? => 0.85€ 9.8 ts? => it2 = 0.415 - g ueres a) t = hang time = ti+t2 = 0.805 (b) Horizontal velocity remains constant vx=0x = ax = 3.18 m/s cc) vertical velocity at take 86 , wing o (d) tano = vy Wys gti= 9.8 (0.34)=3.82 m/s se= tan" (2013) - 50.190

I e) Yialiam Ymen=2.4m Yg = 8.64 m. from similar calculations using @ we get ymax - yi = gti →> t = 0.495 and Yt - Ymax = gta - t2 = 0. 59 s => t = titte = 1.09 s A-10 At max, height itin vertical velocity is zero. » At half the max beight we need to find velocity let max height be H This using roo-zas -0. For max height 0-0 =-2gH for half the max. height - 103 = - 29 H => vž-03 = -dong oslat 1. We are given. x = 0.43 ( speed at half the max height) . 40.43 S Ux' + voete 70.43 Ux+ Vý => Vx+= 60.493 RUx? + výja tano huo = 0.1849 vx' +0,09245 Wy? - tanillohet = 0.113 => post = 0.526 = ton" (2.9692) = 71.380