A fire hose held near the ground shoots water at a speed of 6.38 m/s. The water rises upwards in an arc then falls back to the ground. If the nozzle is pointed upwards at an angle of 56.53degrees, what distance away from the nozzle, in metres, does the water hit the ground?
t(6.38)(sin 56.53)/4.9
=5.32s
d=(6.38)(56.53)
=33.95m
This looks right to me but i wanted to check anyway.
3.3
To find the distance away from the nozzle where the water hits the ground, we can use the following steps:
1. Calculate the time it takes for the water to reach the ground. We can use the vertical motion equation h = ut + (1/2)gt^2, where h = 0 (since it reaches the ground), u is the initial vertical velocity (in this case, it's the vertical component of the initial velocity), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Solving for t, we have:
0 = u*sin(theta)*t - (1/2)gt^2
Rearranging the equation, we get:
t = (2*u*sin(theta))/g
2. Calculate the horizontal distance traveled by the water in this time. We can use the horizontal motion equation d = ut + (1/2)at^2, where d is the horizontal distance traveled, u is the initial horizontal velocity (in this case, it's the horizontal component of the initial velocity), a is the horizontal acceleration (which is 0 since there is no horizontal force acting on the water), and t is the time. As the horizontal acceleration is 0, the equation simplifies to:
d = u*cos(theta)*t
3. Plug in the given values into the equations:
u = 6.38 m/s (initial velocity)
theta = 56.53 degrees (angle of projection)
g = 9.8 m/s^2 (acceleration due to gravity)
4. Calculate the time taken for the water to reach the ground:
t = (2*6.38*sin(56.53))/9.8
t = 5.32 seconds (approximately)
5. Calculate the horizontal distance traveled by the water:
d = 6.38*cos(56.53)*5.32
d = 33.95 meters (approximately)
Therefore, the water hits the ground approximately 33.95 meters away from the nozzle.