A ball is thrown vertically upward …

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A ball is thrown vertically upward direction with a velocity 20m/sfrom the top of the buliding the height of the point from where the ball is thrown is 25m from the ground g=10m/s^2 a) How high the ball will rise? b) How long will it be for the ball hits the ground? Plzzz answer me fast @diti?

Posted by #Aditi~ Angel???? 6 years, 4 months ago

CBSE > Class 11 > Physics

8 answers

#Aditi~ Angel???? 6 years, 4 months ago

Bolo kya hua?

1Thank You

Ayush Vishwakarma?? 6 years, 4 months ago

Are aditi kaha ho tum kaise ho kya hua sister tere ko koi problem hai kya bolo aa ni rhi ho kyn ... plzzzzzzzz aao naaa

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Ayush Vishwakarma?? 6 years, 4 months ago

Kya hua aditi kyn ni aa rahi ho Aajao plzz.,.

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Ayush Vishwakarma?? 6 years, 4 months ago

Sahi hai ya galat meko ni pta but mai try kiya huu thik............ 1) v^2= u^2+2as.,,0= 20^2-20s,,0=400-20s,,then s=20 after that maximum height =20+25= 45m. And.... ............ the second part 2) s = ut+1/2at^2 lagana then quadratic equation banega jisme tera ( TIME =5SEC HOGA)

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Ayush Vishwakarma?? 6 years, 4 months ago

Sahi me mai abhi ni padha huu iseliye hme ni pta chala

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#Aditi~ Angel???? 6 years, 4 months ago

Wrong answer

2Thank You

Ayush Vishwakarma?? 6 years, 4 months ago

Mst banaya hai bhai jgab... .......

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Gaurav Seth 6 years, 4 months ago

(a)
H = u²/(2g)
H = 20²/(2×10)
H = 20 m

The ball will reach 20 m high from the point of projection.
It reaches 20 + 20.5 = 40.5m high from the ground.

(b)
t1 = 2u/g
t1 = 2×20/10
t1 = 4 seconds

If v is the velocity with which it hits the ground then,
v = √(u² + 2aS)
v = √(20² + 2×10×25)
v = 30 m/s

v = u + at
t2 = (v - u)/g
t2 = (30 - 20)/10
t2 = 1 second

Time taken for it to hit the ground = t1 + t2
= 4 + 1
= 5 seconds

Note: I took g = 10m/s².