Answer would be c as shown in the photo
The initial velocity (its speed before falling off) is approximately 1.95 m/s
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Answer:
v_oy = 16.33 m/s
Explanation:
To find the vertical velocity of the tiger, you use the information about the horizontal velocity and maximum horizontal distance traveled.
You use the following formula for the range of the trajectory:
( 1 )
v_ox: horizontal initial velocity = 4.5m/s
v_oy: vertical initial velocity = ?
g: gravitational acceleration = 9.8m/s^2
x_max: range of the trajectory = 15 m
You do v_oy the subject of the formula ( 1 ) and you replace the values of the other parameters in order to calculate v_oy:

hence, the initial vertical velocity of the tiger is 16.33m/s