Question

Find time of flight and range

Answer 1

thanks for questioning

Answer 2

Answer:

Time of Flight = 9 + √241/ 10 s

Horizontal Range, R= 48 m

Explanation:

Given;-

         h = 80 m ( height of pole )

         v = 15 m/s ( velocity of upward projection )

So, finding component of V;

        Vy = u sin ∅

             = 15 × sin 37

             = 15 × 3/5

             = 3 × 3

             = 9 m/s

Also, Vx = u cos ∅

             = 15 ×cos 37

             = 15 × 4/5

             = 3 × 4

             = 12 m/s

Now, finding time of flight;-

Since,    

         S = ut + 1/at²  [Let upward be -ve, and downward be +ve]

       80 = - 9t + 1/2 × 10 × t²

       80 = - 9t + 5t²

        5t² - 9t - 80 = 0

Using Quadratic rule,

a = 5, b = -9 and c= - 80

Discriminant (D)= b² - 4 ac

         D = (-9)² - 4(5)(-80)

         D = 81 + 160

         D = 241

Now, t = - b ± √D/ 2a

         t = 9 ± √241/ 2 × 5

         t = 9 + √241/ 10 or t = 9 - √241/10 s.

         t = 9 + √241/ 10 s

Now, finding range;

        R = u Tf  [where R= horizontal range, u= initial velocity, Tf= time of flight]

        R = 12 × √2 h/ g   [since, Tf = √2h/g]

        R = 12 × √2 ×80/ 10

        R = 12 × √16

        R = 12 × 4

        R = 48 m

Therefore,

        Time of Flight = 9 + √241/ 10 s

        Horizontal Range, R= 48 m