Question

A tiger leaps with a horizontal speed of 4.5 m/s from a boulder and lands 15 meters away.What is the vertical velocity with which the tiger lands?
661 m/s down




32.7 m/s down




2.94 m/s down




0.34 m/s down

Answer 1

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:

$x_{max}=\frac{2v_{ox}v_{oy}}{g}$     ( 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:

$v_{oy}=\frac{gx_{max}}{2v_{ox}}=\frac{(9.8m/s^2)(15m)}{2(4.5m/s)}\\\\v_{oy}=16.33\frac{m}{s}$

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