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Particle a time taken 8.pls

Particle B time taken 2.91

Answer
As "A" projected speed is 30m/s the downward speed of A when it reaches the projected point will also be 30m/s. That means the time taken by A = time taken by B ( As it has same downward speed as A) + time taken by A to reach the projected point. So, First let's calculate the time for B Let's consider g= 10m/s^2 S = ut+ 1/2 gt^2 130 = 30t + 5t^2 t^2 + 6t - 26 = 0 t = (-6+-√36+104)/2 By solving equation, We get t = 2.91 sec The time taken by B = 2.91sec Let's calculate the time for A, The speed of A will become 0 when it reaches it highest point. V = u+ gt 0 = 30 -10t t = 3 sec It will take 3 sec more to reach the projected point So the time taken by A = 3+3+2.91=8.91sec
Answer

We assume g = 10 m/s/s

Now for B going down the equation is s = ut + 1/2 g t^2  Or  130 = 30 t + 5 t^2   Or t^2 + 6t - 26 = 0

Which gives t = -6+- sqrt(36 + 104)/2 = 2.9 sec                  Ans

The other body A will first go up. It will rise say s m till such time its velocity becomes zero. Or 0 = 30^2 -2 x 10 x s

This gives s = 45m We then have 45 = 30 t -1/2 x 10 x t^2  Or t^2 -6t + 9 = 0  This gives (t-3) x (t-3) = 0  Or t = 3

Therefore this body A will first rise for 3 seconds. By symmetry, it will also for 3 seconds and at that point its velocity will again be 30 m/s

Thereafter it will fall just like body B and will take another 2.9 sec.

So the total time taken by B will be 3 + 3 + 2.9 = 8.9 sec               Ans

Answer

So its given both objects are at height = 130 m  the 1st lets look at B we find that , its going down with intial velocity = 30 m/s call this= B0 Lets call final velocity of B be Bf then by Newtons Equation of motion , (Bf^2) - (B0^2) = 2*g*130  , where 130 = distance and g = 9.81 m/s2 = gravitational accl. , so we get here Bf = ((2*9.81*130)+(30^2))^(1/2) = 58.742 m/s then we use Bf = B0 + 9.81*t  , t = time of fall of B so t = 2.93 seconds Then for A; Time for A = Time of going Up above 130 m + Time to fall back to 130 m + Time to reach ground from 130 m As A will have same velocity on falling back to 130m as of B so Time of B = Time to reach ground  from 130 m for A and Time of going up for A  = Time of fall back to 130 m  Thus Time of going up , 0= 30 - 9.81*T = 3.058 sec Total time for A = 2*3.058+2.93 = 9.046 Thus time tken by A  = 9.046 sec and by B = 2.93 sec

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