d Problem 1:An archer shoots an arrow at a 75 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. If d = 75 m and v = 37 m/s. At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 37m/s? | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

# Problem 1:An archer shoots an arrow at a 75 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. If d = 75 m and v = 37 m/s. At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 37m/s?

Problem1 : An archer shoots an arrow at a 75 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. If d = 75 m and v = 37 m/s. At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 37m/s?

Solution:

Here the archer shoots an arrow at some certain angle – it is a projectile motion problem !

So, lets start!

Let the angle is ​$$\theta$$​.

The projectile i-e the arrow is to cover a horizantal distance d= 37 m – which we call its range ‘R’

i-e here we take ​$$d=R=75 m$$​.

Initial velocity of the projectile is V=37/m/s.

We have to find the angle ​$$\theta$$​ at which the projectile (the arrow) should be lauched (thrown).

So, we have the following data;

$R= 75m$

$V=37 \frac{m}{s}$

$\theta=?$

Now we know that range of a projectile launched at angle ​$$\theta$$​with velocity V is given by:

$R=\frac{V^2sin2\theta}{g}$

Where ‘g’ is the gravitational acceleration.

so, putting values of R. V, ​$$\theta$$​ and g in this formula.

$75=\frac{(37)^2sin2\theta}{9.8}$

so, we get

$sin2\theta=\frac {75\times9.8}{37\times37}=0.53$

$2\theta=sin^-1(0.53)=32$

$\theta=\frac{32}{2}=16$

So, the archer should shoot the arrow at 16 degrees angle/