The person's horizontal position is given by
and the time it takes for him to travel 56.6 m is
so your first computed time is the correct one.
The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity 
at time 
is
which tells us that he would reach the ground at about 
. In this time, he would have traveled
But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity 
would have been a bit smaller than 
.
Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;
Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.
f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;
Here, p is the power of the lens.
Put f= 35.71 cm.
p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
Answer:
1. 3 m
2. 27 s
Explanation:
1. "A car traveling at +33 m/s sees a red light and has to stop. If the driver can accelerate at -5.5 m/s², how far does it travel?"
Given:
v₀ = 33 m/s
v = 0 m/s
a = -5.5 m/s²
Unknown: Δx
To determine the equation you need, look for which variable you don't have and aren't solving for. In this case, we aren't given time and aren't solving for time. So look for an equation that doesn't have t in it.
Equation: v² = v₀² + 2aΔx
Substitute and solve:
(0 m/s)² = (33 m/s)² + 2(-5.5 m/s²) Δx
Δx = 3 m
2. "A plane starting from rest at one end of a runway accelerates at 4.8 m/s² for 1800 m. How long did it take to accelerate?"
Given:
v₀ = 0 m/s
a = 4.8 m/s²
Δx = 1800 m
Unknown: t
Equation: Δx = v₀ t + ½ a t²
Substitute and solve:
1800 m = (0 m/s) t + ½ (4.8 m/s²) t²
t ≈ 27 s
