Answer:
tmax = 1.11 s, d = 35.98 m
Explanation:
Here is the complete question
Water leaves a fireman's hose (held near the ground) with an initial velocity
v
0
=
19.5 m/s at an angle θ
= 34 degrees above the horizontal. Assume the water acts as a projectile that moves without air resistance. Use a Cartesian coordinate system with the origin at the hose nozzle position.
Using v
0
,
θ
,
and g, write an expression for the time, t max, the water travels to reach its maximum vertical height. At what horizontal distance d from the building base, should the fireman place the hose for the water to reach its maximum height as it strikes the building? Express this distance, d, in terms of v
0
,
θ
,
and g.
Solution
Since the water leaving the hose is considered to be a projectile motion with initial velocity,v₀ = 19.5 m/s and an angle θ = 34. The time tmax it takes the water to reach maximum height is given by
tmax = v₀sin
θ/g = 19.5 × sin34/9.8 = 19.5 × 0.5592/9.8 = 10.904/9.8 = 1.113 s ≅ 1.11 s
The horizontal distance,d from the base of the building at which the hose must be placed to reach maximum height is the range of the projectile and is given by
d = v₀²sin2θ/g = 19.5²sin(2×34)/9.8 = 19.5²sin68/9.8 = 380.25 × 0.9272/9.8 = 352.562/9.8 = 35.98 m
