Answer to Question #244460 in Physics for Kimberly

Question

4. A Stone is thrown with a horizontal velocity of 20m𝑠 −1 from the top of a cliff 15m high. Air resistance is negligible, for this stone:

(i) Calculate the time to fall 15m,

(ii) Calculate the magnitude of the resultant velocity after falling 15m,

(iii) Describe the difference between the displacement of the stone and the distance that it travels.

Answer 1

(i) $s=gt^2/2\to t=\sqrt{2s/g}=\sqrt{2\cdot15/9.81}=1.75\ (s)$

(ii) $v=\sqrt{(gt)^2+v_h^2}=\sqrt{(9.81\cdot1.75)^2+20^2}\approx26\ (m/s)$

(iii) the displacement of the stone $d=\sqrt{15^2+(20\cdot1.75)^2}=38\ (m)$

the distance of the stone $s=\int_0^{35}\sqrt{1+(9.81x/20^2)^2}\ dx=46\ (m)$