Question

A body P is thrown vertically up with velocity 30ms^-2 and another body Q is thrown up along the same vertically line with same velocity but 1 second later from the ground. When they meet (g=10ms^-2) ​

Answer 1

Answer:

h=ut−  

2

1

​  

gt  

2

 

h  

p

​  

=30×1−  

2

1

​  

×10×1  

2

 

h  

p

​  

=30−5=25m

v=u−gt

v  

p

​  

=30−10×1=20ms  

−1

 

​  

 

Let, t further time they take to meet and meet at height h

$$\begin{array}{l}h = ut - \frac{1}{2}g{t^2}\\h = {h_p} + {u_p}t - \frac{1}{2}g{t^2}\\h = {u_a}t - \frac{1}{2}g{t^2}\\{h_p} + {u_p}t - \frac{1}{2}g{t^2} = {u_a}t - \frac{1}{2}g{t^2}\\{h_p} = \left( {{u_a} - {u_p}} \right)t\\\frac{{{h_p}}}{{{u_a} - {u_p}}} = t\\\frac{{25}}{{30 - 20}} = t\\t = 2.5\,{\rm{secconds}}\\

\end{array}$$

Therefore,

P travels 1+t=3.5