Question

part A is 12.9

so what is Part B and Part C ?

QUESTION 1 I throw a basketball with a speed of 20 m/s at 50 degrees from the ground. At 1.1 seconds after I throw the ball, what is the A) horizontal velocity vx and B) vertical velocity vy? How many seconds after I throw the ball will it have a vertical velocity that is the answer B but negative (ball is going down instead of up)?(round all answers to 1 dp) 12.9

Answer 1

Given u= 20m/s. U 0 - 50 My= usino t = 1.15 Ay=-g for horizontal component 0 / using formula, uz utat, we have, V x = Uyta, t u= uloso axzo Vx = UCOS O + (0 x 1.15) Vx = U COJQ = 20 x Cos 150°) m/s x = 12.9 m/s We have for vertical component using formula .V=urat, Vy = Uy tayt = usino -gt Vy = (20x sin(509 - 9.81X1-1) m/s Vy Vy 4.5 m/s We have now, taking Vy = - 4.5 m/s then using again V=urat, Vy = Uy tagt -4.5 m/s = (20 sin 50- 9.81 t) m/s. 9081+ - (15:32 +405) m/s 19.82 9081 Hence after 2.0s, vertical relocuity of the ball is - 4.5 m/s t = 5 = 2.0 seconds.