A tiger leaps horizontally out of a tree that is 3.70 m high. If she lands 4.80 m from the base of the tree, calculate her initial speed v 0 . (Neglect any effects due to air resistance.)

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

6 cos 34, 6 sin 34 trig

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