Question

A stone is dropped from rest from the top of a cliff. it takes delta t=3.3s for it to reach the ground. parts a-f.

i 7% Part (a) what is the initial velocity, vi, of the stone in ms? Grade Summary Deductions 0% Potential 100% sinO | cos() | tan() | π1CI! -71819 cotan0 asin0 acosO atan0 acotan0 sinh0 cosh0 tanh0cotanh0 4 5 6 Attempts remaining 1 2 3 % per attempt) detailed view END e Degrees Radians BACKSPACE DEL CLEAR Submit Hint Feedba I give up! Hints:0% deduction per hint. Hints remaining: 1 圖 17% Part (b) what is the value of the magnitude of acceleration, in m/s2? 圖 17% Part (d) Calculate the magnitude of vf in ms. Feedback: 0% deduction per feedback. 17% Part(c) Express the final velocity vf of a object travelling with an acceleration a for a time period t which starts with an initial vel tyvi 丛17% Part(e) Express the displacement Δ of an object travelling with acceleration a during a time period ar that has an ni al ve ocity v Δ 17% Part (f) Calculate the magnitude ofAy in meters.

Answer 1

This question is a case of free fall of an object under gravity. Since acceleration due to gravity is constant vector so this is a case of motion under constant acceleration where the three equations of motion are applicable.

This question is a case of free fall of an object under gravity. Since acceleration due to gravity is constant vector so this is a case of motion under constant acceleration where the three equations of motion are applicable. When an object is dropped in air, its initial velocity is zero and its acceleration is constant in magnitude and direction, i e, 9.8 m/s^2 in vertically downward direction. Part rightarrow (a) v_i = 0 m/s part rightarrow (b) a = g = 9.8 m/s^2 part rightarrow (c) v_f = v_i + a delta t part rightarrow (d) v_f = 0 + (- 9.8) (3.3) = -32.34 m/sVertical-bar v_f Vertical-bar = 32.34 m/s part rightarrow (e) delta y = v_i delta t + 1/2 a delta t^2 part rightarrow (f) delta y = 0 (delta t) + 1/2 (- 9.8) (3.3)^2 delta y = -53.361 mVertical-bar delta y Vertical-bar = 53.361 m