A rookie quarterback throws a football with an initial upwar

Step 1
Given;
- ${\displaystyle v_{iy,Chirpy}=0\frac{m}{s}}$
- ${\displaystyle t_{Chirpy}=3.50s}$
- ${\displaystyle v_{i,Milada}=95.0c\frac{m}{s}=0.95\frac{m}{s}}$
- ${\displaystyle {\theta }_{i,Milada}={32.0}^{\circ }}$
${\displaystyle g=9.80\frac{m}{s^2}}$
Step 2
- First of all, we need to find the height of the cliff by using the time the cricket Chirpy takes to reach the ground. So we can use the kinematic formula of vertical displacement.
${\displaystyle y_f=y_i+v_{iy}t+\frac{1}{2}{a}_y{t}^2}$
- We know that the initial vertical velocity component of Chirpy is zero since it jumps horizontally. So,
${\displaystyle y_f=y_i+0t+\frac{1}{2}{a}_y{t}^2}$
- We chose the ground to be the zero level in which ${\displaystyle y=0,so{y}_{f,Ch}=0}$.
${\displaystyle 0=y_i+\frac{1}{2}{a}_y{t}^2}$
Step 3
- We can assume that air resistance is negligible, so the vertical acceleration of Chirpy is the free-fall acceleration, thus ${\displaystyle a_y=-g}$
${\displaystyle 0=y_i-\frac{1}{2}{t}^2}$
- Noting that the initial height of the two crickets is the height of the ckiff, so ${\displaystyle y_i=h}$.
${\displaystyle 0=h-\frac{1}{2}{t}^2}$
- Solving for h;
${\displaystyle h=\frac{1}{2}{t}^2}$
- Plugging the given;
${\displaystyle h=\frac{1}{2}\cdot 9.80\cdot {3.50}^2}$
${\displaystyle h=60.0m}$
Step 4
- Now we know the initial height of the two crickets and we know that the air resistance is neglible.
- This means that the horizontal velocity component of both of the two crickets is constant.
- So, to find the horizontal displacement of Milada, we need to find the time it takes to reach the ground.
Step 5
- We can use the kinematic formula of vertical displacement since it is moving under the free-fall acceleration.
${\displaystyle y_f=y_i+v_{iy}t+\frac{1}{2}{a}_y{t}^2}$
- Notice that ${\displaystyle a_y=-g}$
${\displaystyle y_f=y_i+v_{iy}t-\frac{1}{2}{t}^2}$
- We know that the final position of Malida is the ground in which ${\displaystyle y=0}$.
${\displaystyle 0=y_i+v_{iy}t-\frac{1}{2}{t}^2}$
- We know in step 3, that ${\displaystyle y_i=h}$, so
${\displaystyle 0=h+v_{iy}t-\frac{1}{2}{t}^2}$
Step 6
- We know that the initial vertical velocity component is given by ${\displaystyle v_{iy}=v_i{\sin \theta }_i}$. Thus;