Question

An Estes toy rocket accelerates in a straight line direction from the ground at 31.0 m/s², and at an angle of 51.0° above the horizontal while the engine is burning fuel. The rocket fuel burns out after 4.9 s and the rocket takes a projectile path to the ground.

Find its horizontal distance from the launch point to where it hits the ground.
Answer in meters please.

Answer 1

H point on t = 1180 - 24. se At point A, the rocket reachot after 4.95 Launch = 31x4.gx Coss! 95.6 m/s Vy = 31x4 g x sins - 118.05 m/s. The distance covered by socket in 4.9 sec 1942 4x31x(4.932 = 342.2m yx sins! 322.2 x Sins) 289 22 m Now, We need to find the time when it to hits the ground after the projectile motion then vertical height will be o O= vt + 118.05 + - 4 g + ² = 0 (4.9 t - 118.05) +=0 vertical height = the + 1/2 at 2 4.9 The horizontal velocity will be same as no force is acting in horizontal direction. distance covered during projectile. = V t = 95.6x24.1 = 2303.96m Distance from launch point = 230296 + y cas! 2303.96 + 3722 Cos 51 - 2538.2 m

Thank You !!!