As per the question the arrow was projected into space with an initial velocity of 42 m/s
Hence [tex]v_{o} =42 m/s[/tex]
We have to calculate the maximum height achieved by the arrow and total time taken to come the same point at which it was projected up into the air
Here air resistance is neglected
When the arrow reaches it s maximum height it's velocity becomes zero.
Hence final velocity [tex]v_{f} =0[/tex]
Putting the equation of kinematics we get-
[tex]v^{2} -u^{2} = 2as[/tex]
⇒[tex]v_{f} ^{2} -v_{0} ^{2} = 2gh[/tex]
⇒[tex]0^{2} -42^{2} =2*[-9.8]*h[/tex] [here g is opposite to displacement]
⇒1764=19.6×h
⇒h[tex]=\frac{1764}{19.6}[/tex]
⇒h=90 m
Now we have to calculate the total time of flight.
Now we have to calculate the time of ascent.
From the kinematics we know that [tex]v= u+at[/tex]
⇒0=u+[-g]×t
⇒ [tex]t=\frac{u}{g}[/tex]
we know that time of ascent is equal to time of descent.
hence total time of flight=2×[u/g]
The total time of flight [tex]T=2*\frac{u}{g}[/tex]
=[tex]2*\frac{v_{0} }{g}[/tex]
[tex]=2*\frac{42m/s}{9.8 m/s^2}[/tex]
= 8.571 second.
(a) The maximum height reached by the arrow is 90 m
(b) The expression for the time the arrow is in the air is ( 2vo / g )
Acceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration (m / s²)v = final velocity (m / s)
u = initial velocity (m / s)
t = time taken (s)
d = distance (m)
Let us now tackle the problem!
Given:
vo = u = 42 m/s
Unknown:
a. h = ?
b. t = ?
Solution:
At the maximum height, the speed is 0 ft/s
[tex]v^2 = u^2 - 2gh[/tex]
[tex]0^2 = 42^2 - 2 \times 9.8 \times h[/tex]
[tex]0 = 1764 - 19.6 \times h[/tex]
[tex]h = 1764 \div 19.6[/tex]
[tex]\large {\boxed {h = 90 ~ m} }[/tex]
[tex]h = ut - \frac{1}{2}gt^2[/tex]
[tex]0 = ut - \frac{1}{2}gt^2[/tex]
[tex]ut = \frac{1}{2}gt^2[/tex]
[tex]2ut = gt^2[/tex]
[tex]2u = gt[/tex]
[tex]t = 2u \div g[/tex]
[tex]\large {\boxed {t = \frac{2v_o}{g}} }[/tex]
If vo = 42 m/s , then:
[tex]t = \frac{2(42)}{9.8}[/tex]
[tex]t \approx 8.6 ~ s[/tex]
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate
