projectile motion questions - thank you!?
Hi! Just doing some practice problems and can't seem to derive the right answer. If you could show your steps and the final answer I'd immensely appreciate it! Thanks!
1. A stone is thrown horizontally from the top of a 41 m hill. It strikes the ground at an angle of 11o. With what speed was it thrown?
2. A large cannon is fired from ground level over level ground at an angle of 79o above the horizontal. This initial speed is 993 m/s. Neglecting air resistance, how far will the projectile travel in the horizontal direction before striking the ground?
- gp4rtsLv 71 decade agoFavorite Answer
The angle will be arctan of Vy/Vx, the vertical and horizontal velocities. The vertical velocity Vy = √[2*g*h] (The same as the object dropping straight down--from energy conservation, m*g*h = 0.5*m*v²
Vy = √[2*9.8*41] = 28.35 m/s
Vy/Vx = tan(11º) = 0.1943
Vx = 28.35/0.1943 = 146 m/s The horizontal velocity at impact is the same as the horizontal throw velocity
The equation of motion for vertical motion is
y = V0y*t - 0.5*g*t²
The equation for horizontal motion is x = V0x*t
V0y = V0*sinθ and V0x = V0*cosθ
re-writing the eqs
y = V0*sinθ*t - 0.5*g*t²
x = V0*cosθ*t
The projectile hits the ground when y = 0
0 = V0*sinθ*t - 0.5*g*t²; t = 2*V0*sinθ/g
put this into the second to get the horizontal distance R
R = (V0²/g)*2*sinθ*cosθ
R = 37.7 km
- 3 years ago
the suitable formulation to describe the substitute in place interior the x direction is ?x = v0_x*t + (a million/2)gt^2 The acceleration appearing on the x direction is 0,so xf = v0_x*t changing this into polar coordinates you get xf = (v0cos(?)t) The time it takes for an merchandise in a symmetrical trajectory to hit the floor may well be desperate from this equation: ?y = v0_y*t + (a million/2)gt^2. Set it equivalent to 0 in view which you go with for to hit upon the time it takes whilst it hits flat on the floor 0 = v0sin(?)t - (a million/2)gt^2 gt^2 = 2v0sin(?)t t =2 v0sin(?)/g the gap it takes for an merchandise to realize the different end of the trajectory is consequently v0^2 sin(2?)/g = h fixing for perspective it quite is (a million/2)arcsin(gh/v0^2)