Verified Answer :
A baseball is thrown horizontally at 25 m/s from a cliff 45 m above the level ground and it hits the ground 75 m from the base of the cliff.
First, we will calculate the time the baseball is in the air by analyzing the vertical movement.
This is a uniformly accelerated motion, where the initial vertical velocity (u) is 0, the displacement (s) is 45 m and the acceleration is that of gravity (a = 9.8 m/s²).
We can calculate the time (t) in the air using the following kinematic equation.
\(s = ut + \frac{1}{2} a t^{2} \\\\45m = (0m/s)t + \frac{1}{2} (9.8m/s^{2} ) t^{2} \\\\t = 3.0 s\)
Horizontally, it travels for 3.0 s at a speed of 25 m/s. The horizontal distance traveled is:
\(3.0s \times \frac{25m}{s} = 75 m\)
A baseball is thrown horizontally at 25 m/s from a cliff 45 m above the level ground and it hits the ground 75 m from the base of the cliff.
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Neglecting air resistance, the distance from the base of the cliff where the baseball hit the ground is approximately 75.75m.
Given the data in the question;
- Initially, the baseball was at rest before it was thrown
- Initial velocity; \(u = 0m/s\)
- Final velocity; \(v = 25m/s\)
- Height of the cliff; \(s = 45m\)
First, we determine the time taken. Using the second equation of motion:
\(s = ut + \frac{1}{2}gt^2\)
Where s is the height of the cliff, u is initial velocity, t is time taken, a is acceleration due to gravity( \(9.8m/s^2\))
We substitute in our values
\(45m = 0 + \frac{1}{2}*9.8m/s^2\ *\ t^2\\\\t = \sqrt{ \frac{45m}{4.9m/s^2}}\\\\t = 3.03s\)
Now, to get the horizontal distance, we simply multiply the velocity of the baseball and time taken
Horizontal distance = \(25m/s\ *\ 3.03s\)
Horizontal distance = \(75.75m\)
Therefore, neglecting air resistance, the distance from the base of the cliff where the baseball hit the ground is approximately 75.75m.
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