Answer:
90m
Explanation:
Given that an arrow is shot horizontally from a height of 4.9 m above the ground. The initial speed of the arrow is 45 m/s.
Since the arrow is shot horizontally, the initial vertical velocity will be equal to zero.
Using the second equation of motion,
h = Ut + 1/2gt^2
substitute h = 4.9m and g into the formula
4.9 = 1/2 × 9.8 × t^2
4.9 = 4.9t^2
t^2 = 4.9/ 4.9
t = 1
For horizontal motion g = 0 and the total time T = 2t
T = 2 × 1
T = 2
Substitute T and U into the equation
R = UT
R = 45 × 2
R = 90m
Therefore, the arrow will hit the ground 90m away from the archer.
Answer: 45 meters.
Explanation:
First, let's calculate the time that the arrow needs to hit the ground.
Let's analyze the vertical axis.
The only force acting on the arrow is the gravitational force, then the only acceleration will be the gravitational acceleration.
We can write this as:
a(t) = -g
where = 9.8m/s^2
For the vertical velocity, we can integrate the above equation over the time:
v(t) = -g*t + v0
where t is our variable, the time in seconds, and v0 is the initial vertical velocity.
But we know that the arrow is shot horizontally, then we don't have initial vertical velocity: v0 = 0m/s.
v(t) = -g*t
and the vertical position can be obtained integrating again.
p(t) = -0.5*g*t^2 + p0
Where p0 is the initial position, we know that this is 4.9 meters, then the position equation is:
p(t) = -0.5*9.8m/s^2*t^2 + 4.9m
Now we want to find the value of t where the position is equal to zero, this will be the time that the arrow needs to hit the ground.
0 = -0.5*9.8m/s^2*t^2 + 4.9m
t^2 = 4.9/(0.5*9.8) s^2
t = √( 4.9/(0.5*9.8) s^2 ) = 1s
Now let's analyze the horizontal case.
As we can neglect the air resistance, the horizontal velocity is constant:
v(t) = 45m/s
And the position will be:
V(t) = 45m/s*t + p0
Where in this case p0 is the position of the archer, that we can define as our zero in this axis, then p0 = 0m
Now we know that the vertical displacement of the arrow when it hits the ground will be:
p(1s) = 45m/s*1s = 45m
