Answer to Question #350246 in Physics for john

Question

An object of mass 2 kg is launched at an angle of 30^o above the ground with an initial speed of 40 m/s. Neglecting air resistance , calculate:

i. the kinetic energy of the object when it is launched from the  the ground.

 

ii. the maximum height attained by the object .

 

iii. the speed of the object when it is 12 m above the ground.

 

Answer 1

Given:

$m=2\:\rm kg$

$\theta=30^\circ$

$v_0=40\:\rm m/s$

$g=9.8\:\rm m/s^2$

(i)

$$K_0=\frac{mv_0^2}{2}=\frac{2*40^2}{2}=1600\:\rm J$$

(ii)

$$h_{\max}=\frac{(v_0\sin\theta)^2}{2g}=\frac{(40\sin30^\circ)^2}{2*9.8}=20\:\rm m$$

(iii)

$$v=\sqrt{v_0^2-2gh}=\sqrt{40^2-2*9.8*12}=37\:\rm m/s$$