Calculate the time of flight for a projectile launched at an angle to the horizontal if the velocity is 200 m/s and the angle is 30 degrees

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Calculate the time of flight for a projectile launched at an angle to the horizontal if the velocity is 200 m/s and the angle is 30 degrees

Physics
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If it's fired on a perfectly flat surface, then we can calculate the time of flight by using: \[y = \frac{1}{2}a_yt^2 + v_{o,y}t + y_o\] because the acceleration is constant. The acceleration in the y direction is -9.8 m/s (gravity) y_o = 0, as the initial position was on the ground. We can find the initial velocity in the y direction by decomposing the given velocity vector: |dw:1433992875412:dw| \[V_y = 200 \, \sin(30^o) = 200 * \frac{1}{2} = 100\] Plug in those values into that equation, solve for t. You'll need to use the quadratic equation. You'll get two answers for t. One of them should be 0. The other will be your time of flight.
\[time of flight = (2usin \theta) \div g\] where theta is the angle with the horizontal insert the values to get the answer
Time of flight=(2u sinx)/g Putting values of u, x and g We get, T=(2×200sin30°)/g =20 seconds

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