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The celing of a long hall is 25m high.What is the maximum Horizontal distance that a ball thrown with a spEedof 40m/s can go without hitting the ceiling of the ball?

Rohith , 6 Years ago
Grade 11
anser 3 Answers
janga vidya charan reddy

Last Activity: 6 Years ago

Hello Rohith,
That’s very simple if you know the concept of projectile motion and equations in kinematics.
Just apply;  S=u X rootover(2h/g)                  [Speed=Distance/Time]
Where rootover(2h/g) stands for time of flight;  S for Maximum horizantal distance; u for speed.
Now put down the given values in the equation.
S=40xrootover(2x25/10)
S=40 rootover5.
That’s your answer!
Hope your query is cleared to a large extent.
If you like the answer please approve it.
 
Thank you,
 
Your regards,
Vidya Charan.

Prerna

Last Activity: 6 Years ago

Hi Rohith,
This is the case of a projectile motion 
First of all we have to find out the angle of projection.
Using , h= u^2 (sinQ)^2 / 2g
      here h= 25 m
                g= 10 m/s^2
              Solving this we get  sinQ = square root 5 / 4 
      By  this we can also find out cos Q  by right angle triangle law , which comes out to be = square root 11/ 4
     The range of a projectile = 2 u^2 sinQ cosQ /g
   Putting values  we get range as 20 × square root 55
 
Hope it helps you☺

surendra surendra

Last Activity: 5 Years ago

GIVEN, h=25m ,u=40m/s g=10m/s ^2                                          H=u^2 sin^2theta /2g substitute the values.  You get the theta. Then find the value of R= u^2sin2theta/ g .Then you get the correct answer

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