Question

ID Motion Begin Date: 1/13/2019 00 AM-Dae Date: 12/2019 11:3900 PM End Date: S/10/2019 11:5900 PM 10%) Problem 2. Aball is thrown vertically with velocity wi from a heght of h-47m and lands onthe ground after a tinne interval Δ.The positive ydirection is upward. Refer to the figure > 13% t (a) what is value of the vertical component of acceleration of the free fall motion in meter, per second-quard? Grade Summary sinO cos 4 5 6 O DegreesRadians 을▲ 13% Part (b) Enter an expression forthe vertical component of aldsplacement ofthe motion. Ay, i terms of intial heighth. її&13%Part (e) Enter an expression for&y in terms of Al-e,and the vertical oonpent ofthe initia velocity ofthe motie.h - 13%Part (d) lf the ball starts from rest, what is in moten per second? 13% Part (e) Enter an expressuon for Δ' in terms ofa and Δν, inthe case when the ball starts from rest. 13% Part(f) Calculate the fall time &. in seconds, when the ball starts from rest. 13% Part (g) Returning now to the general case, , 40, ener an expreso n for the vertal om porer t ofthe final vel city,right befre - hall hits the ground, in terms of v,a, and Af 13% Part (h) Calculate the value of wr in meten per second when the ball starts from rest.

Answer 1

(A) a = 9.8 m/s²

(B) what kind of expression the question is asking ..... Not clear....usually, $\Delta$y is h_f - h_i where f is for final and i is for initial.

(C) $\Delta$y = v_i$\Delta$t + 1/2a$\Delta$t²

(D) If ball starts from rest, it means v_i is zero.

(E) $\Delta$y = v_i$\Delta$t + 1/2a$\Delta$t² if ball starts from rest, v_i = 0

$\Delta$y = 1/2a$\Delta$t²

$\Delta$t = sqrt (2$\Delta$y / a)

(F) $\Delta$t = sqrt (2*47 / 9.8)

$\Delta$t = 3.097 sec

(g) v_f = v_i + a$\Delta$t

(h) v_f = 0 + 9.8*3.097

v_f = 30.35 m/s