A stone project horizontally from the top of a towel with a speed of 5m/s. It land in the ground level at a horizontal distance of 20m from the foot of the tower. Calculate the high of the tower
To find the height of the tower, we first need to determine the time it takes for the stone to hit the ground. Since the stone is projected horizontally, its initial vertical velocity is 0 m/s.
We can use the horizontal distance the stone travels and its horizontal velocity to find the time of flight. The formula for distance is:
\[ d = v \times t \]
where \( d \) is the distance (20 m), \( v \) is the horizontal velocity (5 m/s), and \( t \) is the time of flight. We need to solve for \( t \):
\[ t = \frac{d}{v} \]
\[ t = \frac{20 \text{ m}}{5 \text{ m/s}} \]
\[ t = 4 \text{ seconds} \]
Now that we know the time it takes for the stone to hit the ground, we can calculate the height of the tower using the kinematic equation for vertical displacement under constant acceleration due to gravity:
\[ s = ut + \frac{1}{2}at^2 \]
where \( s \) is the vertical displacement (the height of the tower), \( u \) is the initial vertical velocity (0 m/s, since the stone is projected horizontally), \( a \) is the acceleration due to gravity (approximately 9.81 m/s²), and \( t \) is the time before impact. Since \( u \) is 0 m/s, the equation simplifies to:
\[ s = \frac{1}{2}at^2 \]
Plugging in the values, we get:
\[ s = \frac{1}{2} \times 9.81 \text{ m/s}^2 \times (4 \text{ s})^2 \]
\[ s = \frac{1}{2} \times 9.81 \text{ m/s}^2 \times 16 \text{ s}^2 \]
\[ s = 0.5 \times 9.81 \times 16 \]
\[ s = 78.48 \text{ m} \]
Therefore, the height of the tower is approximately 78.48 meters.