boy on the edge of a vertical cliff 20 m high throws a stone horizontally upwards with a speed of 25 m\s it strikes the ground at what horizontal distance from the foot of the cliff use g=10m\s^2
find t with
5t^2 = 20
then use that t to find the distance d in
d = 25t
To find the horizontal distance from the foot of the cliff where the stone strikes the ground, we need to analyze the motion of the stone in the horizontal and vertical directions separately.
First, let's analyze the vertical motion. The stone is thrown upwards, but due to the acceleration of gravity, it will eventually come down and strike the ground. We can use the equation of motion to find the time taken by the stone to reach the ground.
Using the equation of motion for vertical motion:
h = ut + (1/2)gt^2
Where:
h = vertical displacement (20 m)
u = initial vertical velocity (0 m/s since the stone is thrown horizontally)
g = acceleration due to gravity (-10 m/s^2, taking downward as negative)
t = time taken to reach the ground (we need to find this)
Plugging in the known values:
20 = (0)t + (1/2)(-10)t^2
Simplifying the equation:
10t^2 = 20
Dividing both sides by 10:
t^2 = 2
Taking the square root of both sides:
t ≈ √2 or t ≈ -√2
We discard the negative value because time cannot be negative:
t ≈ √2
Now that we have the time taken by the stone to reach the ground, we can find the horizontal distance traveled. Since the stone is thrown horizontally, there is no horizontal acceleration (assuming no air resistance), and the horizontal velocity remains constant throughout the motion.
Using the equation of motion for horizontal motion:
s = ut + (1/2)at^2
Where:
s = horizontal distance (we need to find)
u = initial horizontal velocity (25 m/s)
a = horizontal acceleration (0 m/s^2, since there is no horizontal acceleration)
t = time taken to reach the ground (√2 seconds, which we previously calculated)
Plugging in the known values:
s = (25)(√2) + (1/2)(0)(√2)^2
Simplifying the equation:
s = 25√2
Therefore, the stone strikes the ground at a horizontal distance of approximately 25√2 meters from the foot of the cliff.
To find the horizontal distance from the foot of the cliff where the stone strikes the ground, we will use the equations of motion. Let's break the problem down step by step:
Step 1: Calculate the time taken for the stone to reach the ground.
We know that the vertical motion of the stone is influenced by gravity, and the initial vertical velocity is 0 since the stone is thrown horizontally. The vertical motion can be represented by the equation:
h = ut + (1/2)gt^2
where:
h = vertical displacement (20 m)
u = initial vertical velocity (0 m/s)
g = acceleration due to gravity (10 m/s^2)
t = time taken
Substituting the given values, we can solve for t:
20 = 0 * t + (1/2) * 10 * t^2
20 = 5t^2
t^2 = 4
t = 2 s
So, it takes 2 seconds for the stone to reach the ground.
Step 2: Calculate the horizontal distance covered by the stone.
To find the horizontal distance, we need to use the formula:
d = v * t
where:
d = horizontal distance
v = horizontal velocity (which remains constant throughout the motion)
t = time taken (2 s)
The horizontal velocity can be calculated using the formula:
v = u + at
where:
u = initial horizontal velocity (25 m/s, since the stone is thrown horizontally)
a = horizontal acceleration (0 m/s^2, since no horizontal force is acting)
Substituting the values, we get:
v = 25 + 0 * 2
v = 25 m/s
Now, substituting the values of v and t in the first equation, we have:
d = 25 * 2
d = 50 m
Therefore, the stone strikes the ground at a horizontal distance of 50 meters from the foot of the cliff.