# 1. A fireman 50.0 m away from a burning building directs a stream of water from a ground-level fire...

1. A fireman 50.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of 30.0° above the horizontal. If the speed of the stream as it leaves the hose is 40.0 m/s, at what height will the stream of water strike the building?

2. A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the horizontal. The projectile lands on a hillside 4.00 s later. Neglect air friction.

(a) What is the projectile’s velocity at the highest point of its trajectory?

(b) What is the straight-line distance from where the projectile was launched to where it hits its target?

3. A soccer player kicks a rock horizontally off a 40.0-m-high cliff into a pool of water. If the player hears the sound of the splash 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

(1) usig equation of trajectory for the projectile y = xtanø - $$- gx^2 \over 2u^2 cos^2ø$$ here x = 50m , u = 40m/s and ø = 30degree then height at which water jet will strike the building y = 50tan30 - $$- 10*50^2 \over 2*40^2 cos^230$$ = 50*0.58 - $$25000 \over 2400$$ = 10.42m (2) Here ø = 30degree u = 60m/s , and given time for stoping of projectile is t = 4 sec (a) As the time of flight from the above information is calculated as T = $$2usinø \over g$$ = 6sec. Thus the particle will be at hieghest point at 3sec. hence particle stops after crossing of heighest point Thus at heighest point only horizotal component of velocity will be none zero Ux = ucos30 = 60*0.866 = 51.96m/s (b) The straight line distance is simply the displacement between launch point and stop point which can be calculated as r = $$\sqrt x^2 +y^2$$ where x and y are horizontal and virtical displacement of the projectile calculating x and y x = ucosø *t =51.96 *4 = 207.84m y = usinø*t -(gt^2)/2 = 30*4 - $$10*16 \over 2$$ = 120 -80 = 40m thus r =...