Question

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

1)P travels for 2.5 Sec
2)Q travels for 3.5 Sec
3)P travels for 3.5 Sec
4)Q travels for 1 Sec​

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

Answer 2

Explanation:

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