Question

Q. 9. A stone is thrown horizontally with the velocity 15m/s. Determine the tangential and normal accelerations of the stone in 1 second after it begins to move

Answer 1

I hope you understand right now.

Answer 2

Answer:

Explanation:

Given

initial horizontal velocity u =15 m/s

t= 1 s

after 1 s vertical velocity of stone is given by

$v_y=u_y+a_yt$

$v_y=0+g\times 1$

$v_y=9.8 m/s$

Horizontal velocity will remain same as there is no acceleration in horizontal direction $v_x=15 m/s$

angle made by total velocity with horizontal is given by

$\tan \alpha =\frac{v_y}{v_x}$

$\tan \alpha =\frac{9.8}{15}=0.653$

$\alpha =33.15^{\circ}$

Therefore normal acceleration is given by$g\cos \alpha =9.8\times \cos 33.15=8.17 m/s^2$

Tangential acceleration $=gsin\alpha =9.8\times 0.546=5.35 m/s^2$