Question

Hi can someone please explain how my teacher got the time and how he did step g? :)An archer fires an arrow with a velocity of 42 m/s at an angle of 35 degrees above horizontal?a. What is the horizontal component of its initial velocity?b. What is the vertical component of its initial velocity?C.What is the maximum height attained by the arrow?d. How long does it take the arrow to reach that height?e.What is the total amount of time that it's in the air?f. How far away does it strike the ground?g. What is the horizontal component of its velocity just prior to impact?h. What is the vertical component of its velocity just prior to impact?i. What is the magnitude of its velocity just prior to impact?j. What is the direction of its velocity just prior to impact?

Answer 1

Answer:

a) Horizontal component of its initial velocity = 34.4 m/s

b) Vertical component of its initial velocity = 24.09 m/s

c) The maximum height attained by the arrow = 29.6 m

d) Time taken to reach the maximum height = 2.458 s

e) Total amount of time spent in the air = 4.916 seconds

f) The arrow strikes the ground after travelling an horizontal distance of 169.11 m

g) Horizontal component of its velocity just prior to impact = 34.4 m/s

h) Vertical component of its velocity just prior to impact = -24.09 m/s

i) The magnitude of its velocity just prior to impact = 42 m/s

j) The direction of its velocity just prior to impact = 35°

Step-by-step explanation:

Initial velocity, u = 42 m/s

Angle, θ = 35°

a) Horizontal component of its initial velocity

`\begin{gathered} u_x=u\cos \theta \\ u_x=42\cos 35 \\ u_x=34.4\text{ m/s} \end{gathered}`

Horizontal component of its initial velocity = 34.4 m/s

b) Vertical component of its initial velocity

`\begin{gathered} u_y=u\sin \theta \\ u_y=42\sin 35 \\ u_y=24.09\text{ m/s} \end{gathered}`

Vertical component of its initial velocity = 24.09 m/s

c) Maximum height attained by the arrow

Note that when the arrow goes up, it will surely come down. It will reach the maximum height at a time that is half of the total time that it will spend in the air

`\begin{gathered} \text{If the total time the arrow will spend in air is t} \\ \text{The time it will take to reach the maximum height is t}_{(1)/(2)} \\ u_y=u_0+gt_{(1)/(2)} \\ 24.09_{}=0+9.8t_{(1)/(2)} \\ 24.09_{}=9.8t_{(1)/(2)} \\ t_{(1)/(2)}=(24.09)/(9.8) \\ t_{(1)/(2)}=2.458\text{ s} \end{gathered}`

Maximum height attained by the arrow is calculated below

`\begin{gathered} H=u_yt_{(1)/(2)}+(1)/(2)gt^2 \\ H=24.09(2.458)+(1)/(2)(-9.8)(2.458)^2 \\ H=59.2-29.6 \\ H=29.6\text{ m} \end{gathered}`

The maximum height attained by the arrow = 29.6 m

d) Time taken to reach the maximum height = 2.458 s (As calculated above)

e) Total amount of time spent in the air

The total amount of time spent in the air = Time taken to reach the maximum height + Time taken to fall back to the ground

t = 2.458 s + 2.458 s

t = 4.916 s

Total amount of time spent in the air = 4.916 seconds

f) How far away does it strike the ground?

This will be the horizontal distance travelled by the arrow. Therefore, the horizontal component of the velocity will be used

`\begin{gathered} S=u_xt \\ S=34.4(4.916) \\ S=169.11\text{ m} \end{gathered}`

The arrow strikes the ground after travelling an horizontal distance of 169.11 m

g) Horizontal component of its velocity just prior to impact

The horizontal component of the velocity prior to impact will still remain as 34.4 m/s

h) Vertical component of its velocity just prior to impact

This will be in opposite to the initial velocity when the arrow was fired

Therefore, the vertical component of its velocity just prior to impact = -24.09 m/s

i) The magnitude of its velocity just prior to impact

`\begin{gathered} v^2=u^2_x+u^2_y \\ v^2=34.4^2+(-24.09)^2 \\ v=\sqrt[]{34.4^2+(-24.09)^2} \\ v=42\text{ m/s} \end{gathered}`

The magnitude of its velocity just prior to impact = 42 m/s

j) The direction of its velocity just prior to impact

`\begin{gathered} \theta=\tan ^(-1)(u_y)/(u_x) \\ \theta=\tan ^(-1)(24.09)/(34.4) \\ \theta=\tan ^(-1)0.7 \\ \theta=35^0 \end{gathered}`

The direction of its velocity just prior to impact = 35°