projectiles (1 Viewer)

[maths lover]

i cant seem to get this question

a particle is to be projected with a speed of 25m/s from the floor of a horizontal tunnel of height 20m. taking g as 10ms^-2, find the greatest horizontal range that can be attained in the tunnel.

 

[fortune_cookie]

Range = (v^2/g)sin(2A)

Maximum Height = (v^2 /g )sin^2(A)

Maximum height = (v^2/g)sin^2(A) = 20

So, (v^2/g) = 20/sin^2(A)

Sub into Range formula

R = [20/sin^2(A) ] * sin(2A)

Range is maximum when A=45degrees

Maximum range = 20/ [ sin^2(45) ] * sin(90) = 20/ [1/2] * 1 = 40m

In an exam you have to derive all the above formulas.

 

[maths lover]

Range = (v^2/g)sin(2A)

Maximum Height = (v^2 /g )sin^2(A)

Maximum height = (v^2/g)sin^2(A) = 20

So, (v^2/g) = 20/sin^2(A)

Sub into Range formula

R = [20/sin^2(A) ] * sin(2A)

Range is maximum when A=45degrees

Maximum range = 20/ [ sin^2(45) ] * sin(90) = 20/ [1/2] * 1 = 40m

In an exam you have to derive all the above formulas.

thankyou for your help, but is this the only method of solving?