Question

a. What is the wheel's angular velocity, in rpm, 10 s later? b. How many revolutions does the wheel make during this time? an e 43. Starting from rest, a DVD steadily accelerates to 500 rpm in 1.0s, rotates at this angular speed for 3.0 s, then steadily deceler- ates to a halt in 2.0s. How many revolutions does it make? ie y Problems 44.l A spaceship maneuvering near Planet Zeta is located at 10' km, relative to the planet, and traveling at 9500 m's. It tuns on its thruster engine and accelerates with a (40i -20k) m/s2 for 35 min. What is the spaceship's position when the engine shuts ofl? Give your answer r = (600-400) + 200k) × as a position vector measured in Km 45. Ill A particle moving in the xy-plane has velocity ai t Vo CAlC at r 0. It undergoes acceleration a- bri-e,j, where b and c are constants. Find an expression for the particle's velocity at a later timer 46. II A projectile's horizontal range over level ground is va sin 20lg At what launch angle or angles will the projectile land at half of its maximum possible range? 47, A projectile is launched with speed vo and angle θ. Derive an a. expression for the projectile's maximum height h b. A baseball is hit with a speed of 33.6 mus. Calculate its height and the distance traveled if it is hit at angles of 30.0, 45.0°, and 60.0P 48, l" A projectile is launched from ground level at angle θ and speed CALC vo into a headwind that causes a constant horizontal acceleration of magnitude a opposite the direction of motion. a. Find an expression in terms of a and g for the launch angle that gives maximum range. b. what is the angle for maximum range if a is 0% of g? 49. il A gray kangaroo can bound across level ground with each jump 810 carrying it 10 m from the takeoff point. Typically the kangaroo leaves the ground at a 20° angle. If this is so a. What is its takeoff speed? b. What is its maximum height above the ground?

Please help me with no 48

Answer 1

From kinematic equation V_yf = V_yi + a_yt at highest point, V_yf = 0 0 = (r sin theta) + (-g) t t = V_0 sin theta/g time of flight (t_a) = 2t = 2 V_0 sin theta/g From kinematic equation S = V_yf t_a - 1/2 a t_a^2 s = V_0 cos theta (2 V_0 sin theta/g) - 1/2 a (2 V_0 sin theta/g)^2 S = 2 V_0^2 sin theta cos theta/g - 1/2 a 4 V_0^2 sin^2 theta/theta^2 S = V_0^2 sin 2 theta/g - 2a V_0^2 sin^2 theta/theta^2 For maximum range, dS/d theta = 0, d^2 S/d theta^2 < 0 dS/d theta = 0 V_0^2/g^2 (g (2 cos 2 theta) - a (4 sin theta cos theta)) = 0 2V_0^2/g^2 (g cos 2 theta - a sin 2 theta) = 0 g cos 2 theta = a sin 2 theta tan 2 theta = g/a theta = 1/2 tan^-1 (g/a) B) with a = 10% g = 0.10 g theta = 1/2 tan^-1 (g/0.109) = 42 degree