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Grade: 12th pass

                        

A ball is thrown vertically upwards from the top of a tower with a speed of 100m/s.it strikes the pound near the base of the tower after 25sec . The height of the tower is

4 years ago

Answers : (8)

Vikas TU
12140 Points
							
The ball would reach at top of tower at time,
T = 2u/g = 2*100/10 = 20 s.
that means to cover the height it take 5 seconds.
velocity at the top of tower on returning would be:
v =100 – g*20 = -100 m/s downwards.
Therefore to cover h distance from 2nd eqn. of motion :
h = 100*5 + 0.5*g*25
  = 500 + 125
  = 625 meter.
4 years ago
Ankit Jaiswal
165 Points
							
Hi 
let us suppose the total height of the tower =h
and hence by appling the equation of motion
s=ut+(1/2)*a*t2
by substituting the vaues 
 (–h) = 100*25 + (1/2)(–g)(25)2        (taking above as positive and down as negative)
 (–h) = 2500 – (1/2)*10*625           ( i have taken g as 10 u can take it as 9.8 it depends on the question)
 (–h) = 2500 – 3125
 (–h) = – 625
 so h = 625 m 
4 years ago
Ankit Jaiswal
165 Points
							
on taking g as 9.8 our answer will be 562.5m
 
 
plz approve both if u found them helpful 
 
thanks.....................................................
4 years ago
mmb
19 Points
							As we know that if a object is thrown with velocity v vertically upwards from a height of h m above the ground the time taken by the ball to hit the ground is T = v/g [1+√(1+2gh/v²)]Given v = 100m/sT =25 s. and g we know is 10 m/s²Putting the values in the equation we get 25 = 100/10 [1+√(1+2×10× h/10000)]Equating the values we get,25=10[1+√(1+h/500)]At last we will get 225/100×500=h+500h=1125-500=625Hence the height of the tower is 625 m
						
3 years ago
Prashant kumar sahu
11 Points
							We know thatTime of acent=u/g;100/10=10sec.Time of decent is also u/g=10sec.Total time given is 25sec and we use 10 sec in time of acent then left time is 15 sec.After time of acent the velocity become zero then the apply free fall concept i.eS=ut+1/2×gt2 =0+1/2×10×15×15=1125m.Then the height of tower is=Total distance -time of decent=1125-500=625
						
3 years ago
niharika shukla
19 Points
							First we will find the initial velocity of 1st ball As final velocity v(f)=0a=-g considering upward as positive directionNow frm third equation of motion v^2=u^2+2asu^2=v^2-2as       =2*980 as s=100Then u=√1960=44.27Now for second ball initial velocity =0Frm secnd eqtn of motion ;for first ball S=ut+1/2at^2  =44.27t - 4.9t^2Nd for second ball applying same eqtn.....but taking s as -ve because it is coming downwards-s=-4.9t^2S=4.8t^2But we have to take the distance abv the ground therefore S=100-4.9t^2.......1st eqnS=44.27t-4.9t^2.....2nd eqnEquating 1st nd 2nd we will get t=100/44.27Putting this value of t in eqn 1 100-4.9(100/44.27)^2100-49000/1959.83100-25.00=75 ans
						
3 years ago
Poirei
11 Points
							
Time of ascent = u/g=100/10=10s.
Time of descent after attaining max.ht= Total time – Ascent time=25-10=15s
Initial velocity in the descent motion, u=0 (freely falling after attaining max ht)
Using equations of motion for descent motion, we have 
s=ut+1/2gt^2
=0+1/2*10*15^2=2250/2
=1125 m
Again, ht. attained in the ascent motion is given by
h=(v^2-u^2)/2g
=(0-100^2)/2*(-10) (Ascent motion, g=-10m/s^2)
=500m
Therefore, ht. of the tower = Ht. attained during descent motion -ht. attained during ascent motion
                                         =1125-500 = 625m
2 years ago
Rishi Sharma
askIITians Faculty
614 Points
							Dear Student,
Please find below the solution to your problem.

let us suppose the total height of the tower =h
and hence by appling the equation of motion
s=ut+(1/2)*a*t2
by substituting the vaues
(–h) = 100*25 + (1/2)(–g)(25)2 (taking above as positive and down as negative)
(–h)= 2500 – (1/2)*10*625 ( i have taken g as 10 u can take it as 9.8 it depends on the question)
(–h) = 2500 – 3125
(–h) = – 625
so h = 625 m

Thanks and Regards
2 months ago
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