Question

#No spamming 1) Which angle gives maximum displacement during a projectile motion ??? 2) An object projected with the same speed at two different angles covers the same horizontal range R. If the two times of flight be T1 and T2, prove that R=1/2gT1T2.

Answer 1

Answer:

angle of 45° gives maximum distance

Answer 2

Answer:

1) 45 degree is the angle required in order to get the maximium horizontal displacement also called horizontal range.

why because, when we consider the equation for finding the horizontal range of a projectile motion, i.e, R=  (u²sin2∅)/g

Range(R): The Range of a projectile is the horizontal distance covered (on the x-axis) by the projectile.  

[to derive this equation we need a figure and a few steps]    

Maximum Height (H): this is the maximum height attained by the projectile OR the maximum displacement on the vertical axis(y-axis) covered by the projectile.    

 

im assuming that the 'displacement you have asked for is along the x axis, i.e, maximum/ farthest it has gone[ground]

because if we are to consider its vertical displacement,i.e, how high it went up... if that is the case then maximum height will be attained when initial angle of velocity is 90 degree not 45

but im hoping that the displacement you asked for is along x axis...

also, please refer to the attachment i put in.

after that, read ↓↓  

R=  (u²sin2∅)/g      

now for R (i.e maximum displacement or range)  to be maximum,

the value of  u²sin2∅ (the numerator) should be maximum so u ofcourse should be high but now we are concerned about the angle so

sin∅ should be maximum   [ by the way ∅ is teta--i couldnt find  the right symbol for teta in the type options.I hope ∅ is correct:)]

now, sin function has a maximum value of 1

we know, sin 90= 1

also in our equation it is sin 2∅

so sin 2∅ should be 1

so 2∅ should be 90 i.e ∅ is 45 degree  

Answer

2)

here the initial speed of both projectiles are the same lets take it as u

but the angles are different let them be β and other be Ф

they have the same range give as R

time of flight of 1st one T1 and second T2

        we have equation for the time of flight T

     T = (2 u sin∅)/g

     so T1= (2u sinβ)/g

and T2=(2usinФ)/g  

since range is the same,

1st case range, R=(u²sin2β)/g            

2nd case range, R=(u²sin2Ф)/g

these are equal, this means sin 2β= sin2Ф

this is possible only when β and Ф are complementary angles

two Ranges are equal for two different angles, which happens only when the two throwing angles are  β and (90 - β)    

T1=(2 u sinβ)/g

T2=2 u sin(90-β)

              g

T2= (2 u cosβ)/g

so now product

T1T2=2u²2sinβcosβ/g²

also we have 2sinβcosβ= sin2β

so,

T1T2=2 u²sin2β    --------- eq (1)  

          g       g

also we had

1st case range, R=(u²sin2β)/g

put this into eq (1)

then T1T2= 2R/g

so then 2R=gT1T2

then R= 1/2 * gT1T2

hense prooved..

hope it helps.....