A ball is thrown horizontally from a cliff and hits the ground 4 seconds later 40 meters from the base of the cliff. How high was the cliff (rounded to the nearest tenth of a meter)?

height h is given by

h = h0 - 4.9t^2
so, to find h0, set h=0

0 = h0 - 4.9(16)
h0 = 78.4m

the horizontal distance is irrelevant to finding the cliff height

78.4

To solve this problem, we can use the equations of motion.

First, let's define our known values:
Time, t = 4 seconds
Horizontal distance, d = 40 meters
Acceleration due to gravity, g = 9.8 m/s^2

We can start by using the equation for horizontal motion:

d = v*t

where d is the horizontal distance and v is the horizontal velocity. Since the ball is thrown horizontally, the horizontal velocity is constant, so we can rewrite the equation as:

d = (v_x)(t)

where v_x is the horizontal component of the initial velocity.

Since the ball is thrown horizontally, the vertical component of the initial velocity is zero. The only acceleration acting on the ball is due to gravity, which is purely vertical. Therefore, the vertical motion can be described by the equation:

h = (v_y)(t) + (0.5)(a)(t^2)

where h is the height of the cliff, v_y is the vertical component of the initial velocity, and a is the acceleration due to gravity.

Since the ball is thrown horizontally, the vertical component of the initial velocity is zero. We can rewrite the vertical motion equation as:

h = (0)(t) + (0.5)(g)(t^2)

Simplifying further, we have:

h = 0.5gt^2

Now we can substitute the known values into the equation:

h = 0.5 * 9.8 * (4^2)
h = 0.5 * 9.8 * 16
h = 78.4 meters

Therefore, the height of the cliff is approximately 78.4 meters.

To find the height of the cliff, we can use the equation of motion for vertical motion:

h = h₀ + v₀t + (1/2)at²

where:
h is the final height (unknown)
h₀ is the initial height (from the cliff)
v₀ is the initial vertical velocity (0 since the ball is thrown horizontally)
t is the time (4 seconds)
a is the acceleration due to gravity (-9.8 m/s²)

Since the ball is thrown horizontally, its initial vertical velocity is 0. Therefore, the equation simplifies to:

h = h₀ + (1/2)at²

Now, we need to find h₀, the initial height of the cliff. Since the ball takes 4 seconds to hit the ground, we can use the horizontal motion equation:

d = v₀t

where:
d is the horizontal distance (40 meters)
v₀ is the initial horizontal velocity (unknown)
t is the time (4 seconds)

We can rearrange the equation to solve for v₀:

v₀ = d / t

v₀ = 40 m / 4 s = 10 m/s

Now that we know v₀, we can substitute it into the equation for h:

h = h₀ + (1/2)at²

h = h₀ + (1/2)(-9.8 m/s²)(4 s)² = h₀ - 78.4 m

Simplifying further:

h = h₀ - 78.4 m

Since the ball hits the ground, the final height is 0:

0 = h₀ - 78.4 m

Solving for h₀:

h₀ = 78.4 m

Therefore, the height of the cliff is approximately 78.4 meters.