Question

A student buys a plastic dart gun and tries to find the maximum horizontal range. The student shoots the gun straight up and it takes 3.7 seconds for the dart to return back to the height of the barrel. What is the maximum horizontal range of the gun?

Answer 1

Answer:

The value is $R_{max} = 33.54 \ m$

Explanation:

From the question we are told that

    The total time of flight is  $t = 3.7 \ s$

Generally from kinematic equation

        $v = u - g * \frac{t}{2}$

So v is the velocity at maximum height and the value is  v = 0 m/s

So

       $0 = u - 9.8 * \frac{ 3.7}{2}$

=>   $u = 18.13 \ m/s$

Here u  is the initial velocity of the dart as it leaves that gun  

Gnerally the horizontal range of the dart is mathematically represented as

         $R = \frac{u ^2 sin 2\theta }{g}$

For maximum horizontal range the value of  $\theta = 45^o$        

So

         $R_{max} = \frac{ 18.13 ^2 sin 2(45) }{9.8}$

=>     $R_{max} = 33.54 \ m$