Question

•••HELP ME•••

Question :-In ground to ground projection if range R is related to the time of flight T according to relation R=15/4T^2 then the angle of projection theta with the horizontal direction is?

With explanation ••••​

Answer 1

Answer:

Explanation:

If you’ve seen my answers to projectile motion questions, then you would know that I’m always using one or more Newton’s equations of motion shown below:

S=Vit+12at2 ————--equation 1

Vf=Vi+at —————equation 2

combine equation 1 and equation 2 to eliminate “t” gives

V2f−V2i=2aS —————equation 3

It is highly recommended to watch your signs in these equations. Velocities are up = positive, down = negative and the acceleration due to gravity always points down so a=−9.81ms2

Since your question involves time of flight, we can use equations 1 or 2.

Let’s write equation 1 with subscripts for motion in the y-direction:

Sy=(Vi)yt+12ayt2

In this form of the equation, distance is measured from the starting point. Remember our sign convention. For the total time of flight, the object goes up and comes back to ground level, so Sy=0

Also, (vi)y=Vsinθ giving

0=(Vsinθ)t+12(−9.81ms2)t2

cancelling “t” gives

0=(Vsinθ)+12(−9.81ms2)t

solving for V gives:

V=9.812sinθt ————— equation 4

Note that using equation 2 would give the same result:

(Vf)y=(Vi)y+ayt —————equation 2

At the highest point (Vf)y=0, and the time to reach the highest point is t2:

0=Vsinθ+(−9.81)t2

or

V=9.812sinθt

To calculate the range, we also usually use equation 2 written for motion in the x-direction:

Sx=(Vi)xt+12axt2

but ax=0

∴ Sx=(Vi)xt

or

Sx=(Vcosθ)t

Your problem states that the range R=Sx=5t2

∴ 5t2=(Vcosθ)t ————- equation 5

substitute equation 4 into equation 5:

5t2=(9.812sinθt)(cosθ)t

or

5=(9.812sinθ)cosθ

or

10sinθ=9.81cosθ

or

sinθcosθ=9.8110

or

tanθ=0.981  

θ=44.5∘

I’m guessing that the question maker probably assumed you were using g=10ms2 (rather than 9.81) which would give the nice number:

θ=45∘