A) 8 m/s
B) 6 m/s
C) 10 m/s
D) data is not sufficient
Correct Answer: C
Solution :
Key Idea : Velocity is the rate of change of displacement. Velocity of particle, v = rate of change of displacement i. e. , \[v=\frac{dr}{dt}\] Given, \[x=6\,6t\] ... (i) and \[y=8\,t-5\,{{t}^{2}}\] ... (ii) Differentiating Eqs.(i) and (ii), we get \[{{v}_{x}}=\frac{dx}{dt}=\frac{d}{dt}\,(6\,t)=6\,m/s\] and \[{{v}_{y}}=\frac{dy}{dt}=\frac{d}{dt}\,(8\,t-5{{t}^{2}})=(8-10\,t)m/s\] At \[t=0\], \[{{v}_{x}}\left| _{t\,=\,0}=6\,m/s \right.\] and \[{{v}_{y}}\left| _{t\,=\,0}=8\,m/s \right.\] Hence, velocity of projection at time \[t=0\] is \[v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}=\sqrt{{{(6)}^{2}}+{{(8)}^{2}}}\] \[=\sqrt{36+64}=\sqrt{100}=10\,m/s\]
You need to login to perform this action.
You will be redirected in
3 sec
