Question

while standing on a cliff a boy tosses a stone into a river below. If he throws the ball horizontally with a velocity of 3.0 m/s and it strikes the water 4.5 m away, how high above the water is the boy

Answer 1

Given data:

* The horizontal velocity of the stone is 3 m/s.

* The range of the stone is 4.5 m.

* The acceleration of the stone in the horizontal direction is zero.

Solution:

By the kinematics equation, the horizontal motion of the stone in terms of range and time is,

$R=u_xt+\frac{1}{2}a_xt^2_{}$

where u_x is the horizontal velocity, a_x is the horizontal acceleration, t is the time at which the stone strikes the water, and R is the horizontal range,

Substituting the known values,

$\begin{gathered} 4.5=3\times t+0 \\ t=\frac{4.5}{3} \\ t=1.5\text{ s} \end{gathered}$

By the kinematics equation, the vertical motion of the stone in terms of the time is,

$H=u_yt+\frac{1}{2}gt^2_{}$

where u_y is the vertical velocity of the stone, g is the acceleration due to gravity, t is the time take to strike the water and H is the height of the stone,

$\begin{gathered} H=0+\frac{1}{2}\times9.8\times(1.5)^2 \\ H=11.025\text{ m} \end{gathered}$

Thus, the height of the cliff is 11.025 m.