a child standing at the edge of a cliff 100m high throws a stone vertically downwards. The stone hits the sea 4s later. Ignoring air resistance, find the initial speed with which the stone is thrown

To find the initial speed with which the stone is thrown, we can use the kinematic equation for vertical motion:

\[ h = v_i t + \frac{1}{2} g t^2 \]

where:
- \( h \) is the height of the cliff (100m),
- \( v_i \) is the initial velocity (what we need to find),
- \( t \) is the time of flight (4s),
- \( g \) is the acceleration due to gravity (-9.8m/s²) in the downward direction.

We can rearrange the equation to solve for \( v_i \):

\[ v_i = \frac{h - \frac{1}{2} g t^2}{t} \]

Plugging in the given values:

\[ v_i = \frac{100m - \frac{1}{2} \times 9.8m/s^2 \times (4s)^2}{4s} \]

Let's calculate it step by step:

1. Find the term inside the parentheses:
\[ \frac{1}{2} \times 9.8m/s^2 \times (4s)^2 = 78.4m \]

2. Subtract this term from the height of the cliff:
\[ 100m - 78.4m = 21.6m \]

3. Divide the remaining term by the time of flight:
\[ \frac{21.6m}{4s} = 5.4m/s \]

Therefore, the initial speed with which the stone is thrown is 5.4 m/s.

To find the initial speed with which the stone is thrown, we can use the equation of motion for vertical motion under gravity:

y = ut + (1/2)gt^2

Where:
- y is the displacement (change in height) = 100m (height of the cliff)
- u is the initial velocity (speed with which the stone is thrown) = ?
- t is the time taken for the stone to hit the sea = 4s
- g is the acceleration due to gravity = 9.8 m/s^2

Now, rearranging the equation to solve for u:

y = ut + (1/2)gt^2
100 = u(4) + (1/2)(9.8)(4)^2
100 = 4u + (1/2)(9.8)(16)
100 = 4u + 78.4
4u = 100 - 78.4
4u = 21.6
u = 21.6 / 4
u = 5.4 m/s

Therefore, the initial speed with which the stone is thrown vertically downwards is 5.4 m/s.

h = -4.9t^2 + vt + 100, where v is the initial velocity

when t = 4, h = 0
0 = -4.9(4) + 4v + 100
4v = -80.4
v = -20.1

the stone was tossed with a velocity of 20.1 m downwards