This question can be solved by using the equations of motion.
a) The initial speed of the arrow is was "9.81 m/s".
b) It took the arrow "1.13 s" to reach a height of 17.5 m.
a)
We will use the second equation of motion to find out the initial speed of the arrow.
[tex]h= v_it + \frac{1}{2}gt^2\\[/tex]
where,
vi = initial speed = ?
h = height = 35 m
t = time interval = 2 s
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}[/tex]
vi = Â 9.81 m/s
b)
To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.
[tex]h= v_it + \frac{1}{2}gt^2\\[/tex]
where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 17.5 m
vi = initial speed = 9.81 m/s
t = time = ?
Therefore,
[tex]17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0[/tex]
solving this quadratic equation using the quadratic formula, we get:
t = -3.13 s (OR) t = 1.13 s
Since time can not have a negative value.
Therefore,
t = 1.13 s
Learn more about equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.
