Crickets chirpy and milada jump from the top of a vertical cliff. chirpy drops downward and reaches the ground in 2.70 s, while milada jumps horizontally with an initial speed of 95.0 cm> s. how far from the base of the cliff will milada hit the ground? ignore air resistance.

The individual directions of motion of Chirpy and Milada are components of projectile motion. A projectile motion is characterized by a motion in the shape of an arc. The thing about projectile motion is, the horizontal component and the vertical component are independent of each other. The horizontal motion acts on constant velocity, while the vertical motion acts on a constant acceleration equal to the force of gravity, 9.81 m/s². Even though they are independent, there is a relationship between them called the trajectory of a projectile equation:

y = xtanθ + gx²/2v²cos²θ
where
y is the vertical height
x is the horizontal range
θ is the angle of inclination
g is 9.81 m/s²
v is the initial velocity

Since Milada jumps horizontally, there is no angle of inclination: θ=0°. The initial velocity is equal to 95 cm/s or 0.95 m/s. Now, we have to determine x. But we can't do that without finding y first. This can be obtained from Chirpy's downward motion. In a free falling motion, the time of flight is equal to

t = √2y/g
2.70 = √2y/9.81
y = 35.757 m

Now, we can solve for x. I suggest you use your scientific calculator so that you can easily solve for x.

35.757 = xtan0° + (9.81)x²/2(0.95)²cos²0°
x = 2.565

Therefore, Melinda hits the ground 2.565 meters away from the base of the cliff.

Where is the table ? you can take a picture of it