Question

a body p is thrown vertically up with velocity 30 m/s and another body q is thrown up along the same vertically line with same velocity but 1sec later from the ground when they meet time​

Answer 1

Answer:

this is the answer

Explanation:

hope you get it

Answer 2

Answer:

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

Explanation:

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