Question

solve this question of physics,(28) if known rightly then answer .... please no wrong solution..thanku​

Answer 1

Answer :-

Applying conservation of energy to the top of the track (D)and point A (as shown in the diagram)

$\sf{(PE)_D+(KE)_D=(PE)_A+(KE)_A}$

$\sf{\therefore mg(2.4)+0=mg(1.0)+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2 }\\\\\sf{=\dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{2}{5}MR^2 )\dfrac{v^2}{R^2}}\\\\\sf{=\dfrac{7}{2}mv^2}$

$\sf{\implies mg \times 1.4=\dfrac{7}{2}mv^2}\\\\\sf{\implies \dfrac{1.4\times9.8\times10}{7}=v^2 }\\\\\sf{\implies v^2=19.6}\\\\\sf{\implies v=\sqrt{19.6}}\\\\\sf{\implies v=4.43\ m/s}$

After point A, the body takes a parabolic path. The vertical motion parameters of parabolic motion will be

$\sf{u_y=0}$

$\sf{S_y=1\ m}$

$\sf{a_y=9.8\ m/s^2}$

$\sf{t_y=?}$

$\sf{S_y=u_yt_y+\dfrac{1}{2}a_yt_y^2 }\\\\\displaystyle{\sf{\implies 1=0\times t_y+\frac{1}{2}\times 9.8 \times t_y^2}}\\\\\displaystyle{\sf{\implies 1=0+4.9t_y^2}}\\\\\displaystyle{\sf{\implies t_y^2=\frac{1}{4.9}}}\\\\\displaystyle{\sf{\implies t_y=\sqrt{\frac{1}{4.9} } }}$

$\sf{\implies t_y=0.45\ secs}$

Applying this time in horizontal motion of parabolic path,

BC = 4.43 × 0.45 = 2 m

Answer 2

twilight queen bee answer is 2m