While standing on an open bed of a truck moving at 35 m/s, and archer sees a duck flying directly overhead. The archer shoots an arrow at the duck and misses. The arrow leaves the bow with a vertical velocity of 98 m/s.
a) how long does it remain in the air
b) the truck maintains a constant speed of 35 m/s and does not change it's direction. Where does the arrow finally land?
c) what horizontal distance does the arrow travel while its in the air?
Vi = 98
v = Vi - 9.8 t
0 at top
98 = 9.8 t
t = 10 seconds to top (good archer, strong bow)
so
a) 20 seconds up and down
b) did that, archer dead in truck bed. always turn after dropping a bomb out of your airplane by the way.
c) 20 seconds at 35 m/s = 700 meters
you are, of course, assuming that there was no horizontal component to the arrow's velocity.
In Heinlein's "Stranger in a Strange Land" there is a character who is a professional observer. If you point to a house on a hill and ask what color it is, he'll reply, "This side is white," making no assumptions about the other sides.
However, given the usual sloppy problems posted here, I agree with your analysis (ignoring air resistance).
This dead archer was a dead shot and the duck was directly overhead :) By the way it was an expert seagull pulling out of a bombing run anyway and the archer was simply retaliating.
When standing on an open truck at a velacity of 35m/s a man sees a duck flying directly overhead the man shorts an arrow at the duck and misses it the arrow leaves the bow with the vertical velocity of 98m/s the truck accelerates to a constant speed of 40m/s in the same direction just after the man has not at the duck if the truck open board at which the man is standing is 2m above the ground . i;how long will the arrow remain in the air before hitting the ground ? ii;where will the arrow land in relation to the position of the truck?
To solve this problem, we need to consider both the horizontal and vertical components of motion. Let's address each part of the problem:
a) How long does the arrow remain in the air?
To find the time of flight for the arrow, we need to consider its vertical motion. We can use the equation for vertical motion:
vf = vo + gt,
where vf is the final vertical velocity, vo is the initial vertical velocity, g is the acceleration due to gravity (approximately -9.8 m/s^2), and t is the time.
In this case, the arrow leaves the bow with a vertical velocity of 98 m/s and eventually reaches zero velocity at the highest point of its trajectory. Therefore, vo = 98 m/s and vf = 0.
0 = 98 - 9.8t,
Simplifying the equation gives us:
9.8t = 98,
Divide both sides by 9.8:
t = 10 seconds.
Therefore, the arrow remains in the air for 10 seconds.
b) Where does the arrow finally land?
Since the truck is moving at a constant speed and does not change direction, we can find the horizontal distance the arrow travels by multiplying the truck's speed by the time of flight.
Horizontal distance = truck speed × time of flight,
Horizontal distance = 35 m/s × 10 s.
The arrow finally lands 350 meters away from the archer's initial position.
c) What horizontal distance does the arrow travel while it's in the air?
During the time of flight, the arrow travels with a constant horizontal velocity determined by the truck's speed, which remains at 35 m/s.
Therefore, the horizontal distance the arrow travels while it's in the air is equal to the product of its horizontal velocity and the time of flight:
Horizontal distance = horizontal velocity × time of flight,
Horizontal distance = 35 m/s × 10 s.
The arrow travels a horizontal distance of 350 meters.
Well first of all, the arrow hits the archer on the way down. Both the truck and the archer have a constant horizontal velocity component of 35 m/s. Your late archer deserved it no doubt
now the vertical problem.