Answer:
Explanation:
Given that
Height of the tree is 3.7m
Therefore yo=3.7m
yox= 0m
The tiger lands 4.8m from the tree
Then, Range x=4.8m
Since she flies horizontally and she lands away from the bottom of the tree, her path will be trajectory as shown in the attachment
Let know the time of flight, using the equation of motion
voy is the initial velocity of the vertical motion of the is the tiger, which is zero at the beginning.
y = y0 + voy*t + ½*g*t²
Given that,
y=3.7m
yo=0m
voy=0m/s
g=9.81m/s²
y = y0 + voy*t + ½*g*t²
3.7=0+0•t+½×9.81×t²
3.7=0+0+4.905t²
3.7=4.905t²
t²=3.7/4.905
t²=0.7543
t=√0.7543
t=0.87sec
Time to reach the ground is
Now to know the initial velocity of the horizontal motion, using equation of motion
x=xo+Voxt
Vox is the horizontal initial velocity of the tiger.
x=xo+Voxt
4.8=0+Vox×0.87
4.8=0+0.87Vox.
4.8=0.87Vox.
Then, Vox=4.8/0.87
Vox=5.52m/s
Then, her initial velocity Vo
Vo=√voy²+vox²
Vo=√0²+5.52²
Vo=√5.52²
Vo=5.52m/s
The initial velocity is 5.52m/s
