An arrow is shot at an angle of 35° and a velocity of 50 m/s. How long does it take to return to its original starting height?

4.02 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 35°

Initial velocity (u) = 50 m/s

Acceleration due to gravity (g) = 10 m/s²

Time of flight (T) =?

The time of flight of the arrow can be obtained as follow:

T = 2uSineθ / g

T = 2 × 35 × Sine 35 / 10

T = 70 × 0.5736 / 10

T = 7 × 0.5736

T = 4.02 s

Therefore, the time taken for the arrow to return is 4.02 s

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