Answer:
0.7 m.
Explanation:
We are given that
Initial velocity of fox,v=12 m/s
[tex]\theta=18.9^{\circ}[/tex]
Distance of fox from the ball=3.2 m
Horizontal component of velocity=[tex]v_x=vcos\theta=12\times cos18.9=11.35m/s[/tex]
Vertical component of velocity=[tex]v_y=vsin\theta=12sin18.9=3.89m/s[/tex]
Time taken by fox to travel in horizontal direction=[tex]\frac{distance}{horizontal\;velocity}=\frac{3.2}{11.35}=0.28 s[/tex]
Time taken by box to the wall=0.28 s
Formula:
[tex]s=vsin\theta t-\frac{1}{2}gt^2[/tex]
Where [tex]g=9.8m/s^2[/tex]
Using the formula
[tex]s=3.89(0.28)-\frac{1}{2}(9.8)(0.28)^2[/tex]
Takine g negative because fox going against to gravity.
[tex]s=0.7 m[/tex]
Hence, the fox will reach upto height 0.7 m when it reaches the wall
Answer:
s = 0.733 m
Explanation:
given,
horizontal distance = 3.5 m
height of the wall,h = 0.5 m
initial velocity, v= 12 m/s
angle with horizontal, θ = 18.9°
Maximum height fox will reach = ?
horizontal velocity of the fox = 12 cos 18.9° = 11.35 m/s
time taken to cover horizontal distance
[tex]t = \dfrac{h}{v_x}[/tex]
[tex]t = \dfrac{3.5}{11.35}[/tex]
t = 0.308 s
vertical velocity = 12 sin 18.9°
= 3.89 m/s
vertical distance travel by box
[tex]s = u t + \dfrac{1}{2}gt^2[/tex]
[tex]s =3.89\times 0.308 - \dfrac{1}{2}\times 9.8 \times 0.308^2[/tex]
s = 0.733 m
Hence, the vertical height travel by the fox is equal to s = 0.733 m
