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VLD [36.1K]
2 years ago
8

You fire a projectile 35° above horizontal with an initial velocity of 200m/s. It lands in a valley 300m below the launch point.

What is the time of flight of the projectile, and what is the range of projectile? ​
Physics
1 answer:
Brilliant_brown [7]2 years ago
3 0

Answer:

See below

Explanation:

Vertical component of initial velocity  =  200 sin 35° = 114.72 m/s

then use position formula      a = 9.81 m/s^2

  0  = 300 + 114.72 t  -  1/2 (9.81)(t^2)

            use quadratic formula with a = - 4.905    b = 114.72   c = 300

              to find t = <u>25.76 seconds </u>

<u />

To find the range

   ( horizontal distance the projectile lands from launch point)

Horizontal component of initial velocity    200 cos 35 = 163.83 m/s

 ( it flies horizontally at this speed for  <u>25.76 seconds <==found above</u>)

    163.83  m/s  *  25.76 s = <u>4220.3 meters</u>

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