An object R leaves a platform XY with a horizontal velocity of 7m-1 and lands at Q. lf it takes the same object 0.3s to fall freely from Y to P, calculate t...

An object R leaves a platform XY with a horizontal velocity of 7m^-1 and lands at Q. lf it takes the same object 0.3s to fall freely from Y to P, calculate the distance PQ.(Take g = 1Oms²)

Answer Details

The time of flight of an object in projectile motion can be calculated using the formula: t = 2u sin?/g where u is the initial velocity of the object, ? is the angle of projection, and g is the acceleration due to gravity. In this question, the object R leaves the platform with a horizontal velocity of 7m/s, which means its initial vertical velocity is 0. We can therefore use the formula: t = 2h/g where h is the height of the platform XY. We can rearrange this equation to solve for h: h = 1/2gt^2 We are given that it takes 0.3s for the object to fall freely from Y to P, so: h = 1/2 × 10 × (0.3)² = 0.45m Now, we can use the fact that the horizontal distance traveled by the object is given by: d = ut where d is the horizontal distance, u is the initial horizontal velocity, and t is the time of flight. Since there is no horizontal acceleration, the time of flight is the same as the time it takes for the object to fall from Y to P. Therefore: d = 7 × 0.3 = 2.1m Finally, we can use Pythagoras' theorem to find the distance PQ: PQ = ?(d² + h²) = ?(2.1² + 0.45²) ? 2.16m Therefore, the answer is 2.10m.