Need someone who's really good at physics? (projectile motion qn)?

Question

Water is ejected from a water canon 45 degrees to the horizontal. The horizontal displacement of the water is 20 metres. Diameter of the circular jet is 2 cm. Density of water is 1000 kg/m^-3. Show that initial speed of the water leaving the canon is 14m/s

Got it! Thanks a million David. You're one smart dude.

Answer 1

The diameter of the jet and the density of water are just distractions; you don't need that information. With a 45 degree angle, the initial horizontal and vertical speeds are equal. The horizontal speed remains constant, but the vertical speed is affected by gravity. The water remains airborn for a time dependent on the initial vertical speed, which I'll call v. The amount of time it takes for gravity to slow the water to a vertical standstill is v/g. It takes an equal amount of time for the water to get back to the level of the ground. In a time of 2v/g, the water travels a horizontal distance of 20 m. The horizontal velocity is also v, so the horizontal distance traveled is 2v^2/g. So 2v^2/g = 20, and v = sqrt(20g/2) = sqrt(10*9.8) = 9.9 m/s. But that's not the total speed. It's the horizontal component, which is also the vertical component, so the actual speed is sqrt(2*9.9^2) = 14 m/s.

Answer 2

20m is the horizontal displacement. let the time be t. so vcos45=20/t or v=20 root 2/t. Now vertical displacement has taken time = t/2 so vsin45 = 10 t/2 assuming g = 10m/s squared. so v = 5 t root 2 . v = v . or 20 root 2 /t = 5 t root 2 therefore t = 2. there fore v = 20 root 2/2 = 10 root 2 = 10*1.414= 14.14 m/s. i hope it isn't a messy explaination you don,t need the density of water or the diameter.

Answer 3

Won't happen.