A projectile is thrown from a point O on the ground at an angle 45° from the vertical and with a speed 52 m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity g=10 m/s2.
The value of x is ____.
Answer: 7.5
Solution
Range without splitting: R=g2uxuy=102×5×5=5 m.
At highest point, the projectile has traveled R/2 = 2.5 m horizontally.
After splitting, the second part has horizontal velocity v = 10 m/s (from momentum conservation) and travels for t = 0.5 s.
Additional horizontal distance = v×t=10×0.5=5 m.
Total distance from O: x=2R+5=2.5+5=7.5 m.
JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Numerical answer
