Question 1

Multiple choiceMedium1 mark

A small ball is launched from a platform. Air resistance is negligible and g=10 m s−2g=10\ \text{m s}^{-2} The figure shows the launch conditions. What is the horizontal distance from the point directly below the launch point to where the ball first reaches the ground?

10 m10 m s-130°
Show answer and explanation

Answer: B. Taking upward as positive, vertical motion gives 0=10+5t−5t20=10+5t-5t^2 so the positive time to reach the ground is t=2.0 st=2.0\ \text{s} The horizontal speed is 10cos⁡30∘ m s−110\cos30^\circ\ \text{m s}^{-1} giving a horizontal distance of (10cos⁡30∘)(2.0)=17.3 m(10\cos30^\circ)(2.0)=17.3\ \text{m}

A: uses the vertical component, 10sin⁡30∘10\sin30^\circ as the horizontal speed.
C: uses the launch speed as the horizontal speed, omitting the cosine component.
D: doubles the flight time after finding it, adding an extra descent interval.

Remember: horizontal and vertical components determine different parts of projectile motion.

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