MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Second Law of Motion

A particle is...

Question

A particle is projected vertically upwards with velocity 20 m/s at t = 0. At what time, particle is at height of 15 m and it is coming down [g = 10 m/s²]

एक कण को t = 0 पर 20 m/s के वेग से ऊर्ध्वाधर ऊपर की ओर प्रक्षेपित किया जाता है। जब कण नीचे की ओर आ रहा है, तब कण किस समय पर 15 m की ऊँचाई पर होगा? [g = 10 m/s²]

A

2 seconds

2 सेकण्ड

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

1 second

1 सेकण्ड

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

2.5 seconds

2.5 सेकण्ड

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

3 seconds

3 सेकण्ड

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 3 seconds

3 सेकण्ड

Solution

Suggest Corrections

thumbs-up

0

Similar questions

Q. A particle is projected vertically upwards with velocity 20 m/s. It will be at height 15 m at two time instant while going up and while coming down. Difference in these two time instants is [g = 10 m/s²]

एक कण को 20 m/s के वेग से ऊर्ध्वाधर रूप से ऊपर की ओर प्रक्षेपित किया जाता है। यह ऊपर जाने के दौरान तथा नीचे आने के दौरान दो समय क्षणों पर 15 m की ऊँचाई पर होगा। इन दो समय क्षणों में अन्तर है [g = 10 m/s²]

Q. Two particles A and B are at vertical separation of 20 m. At t = 0 particle A is projected vertically upwards with velocity 20 m/s whereas B is just dropped along same line of motion as shown. At what height above the ground particles will collide? [g = 10 m/s²]

दो कण A तथा B, 20 m की ऊर्ध्वाधर दूरी पर हैं। t = 0 पर कण A को 20 m/s के वेग से ऊर्ध्वाधर रूप से ऊपर की ओर प्रक्षेपित किया जाता है, जबकि B को ठीक इसी समय पर दर्शाए अनुसार गति की समान रेखा के अनुदिश गिराया जाता है। धरातल से कितनी ऊँचाई पर कण एक दूसरे से टकराऐंगे? [g = 10 m/s²]

Q. Two particles are projected from ground with a speed of

40 m/s

at complimentary angles. If the difference in the maximum height of the particles is

50 m

, then the height of the individual particles is

What will be the area A of the quadrilateral in each case with their diagonal $d c m$ & the sum of their heights $(h_{1} + h_{2}) c m$ .

i) $d = 20, (h_{1} + h_{2}) = 10$

ii) $d = 15, (h_{1} + h_{2}) = 20$

iii) $d = 20, (h_{1} + h_{2}) = 25$

iv) $d = 40, (h_{1} + h_{2}) = 30$

The correct statement with regard to $H_{2}^{+}$ & $H_{2}^{-}$ is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Second Law of Motion

PHYSICS

Watch in App

explore_more_icon

Explore more

Second Law of Motion

Standard IX Physics

Join BYJU'S Learning Program

CrossIcon