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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
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

```							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
```							Hi let us suppose the total height of the tower =hand hence by appling the equation of motions=ut+(1/2)*a*t2by 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 =  m
```
4 years ago
```							on taking g as 9.8 our answer will be 562.5m  plz approve both if u found them helpful  thanks.....................................................
```
4 years ago
```							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
```							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
```							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
```							Time of ascent = u/g=100/10=10s.Time of descent after attaining max.ht= Total time – Ascent time=25-10=15sInitial 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 mAgain, ht. attained in the ascent motion is given byh=(v^2-u^2)/2g=(0-100^2)/2*(-10) (Ascent motion, g=-10m/s^2)=500mTherefore, ht. of the tower = Ht. attained during descent motion -ht. attained during ascent motion                                         =1125-500 = 625m
```
2 years ago
```							Dear Student,Please find below the solution to your problem.let us suppose the total height of the tower =hand hence by appling the equation of motions=ut+(1/2)*a*t2by 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) = – 625so h = 625 mThanks and Regards
```
6 months ago
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