a physics book is thrown horizontal at a velocity of 10.0m/s from the top of a cliff 78.4 m high. a). how long does the stone take to reach the bottom of the cliff? b.) how far from the base of the base of the cliff does the stone stike the ground?
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To tackle this problem, we can break it down into two parts: the horizontal motion and the vertical motion of the stone.
a) Let's first analyze the vertical motion to find the time it takes for the stone to reach the bottom of the cliff. We can use the equation of motion in the vertical direction:
y = y0 + v0y*t - (1/2)*g*t^2
Where:
- y is the final vertical displacement (78.4 m)
- y0 is the initial vertical displacement (0 m)
- v0y is the initial vertical velocity (0 m/s as the stone is thrown horizontally)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time
Plugging in the values into the equation, we get:
78.4 = 0 + 0 - (1/2)*9.8*t^2
Rearranging the equation, we get:
4.9*t^2 = 78.4
Dividing both sides by 4.9, we have:
t^2 = 16
Taking the square root of both sides, we find:
t = 4 seconds
Therefore, it takes 4 seconds for the stone to reach the bottom of the cliff.
b) Now, let's analyze the horizontal motion of the stone. Since there are no horizontal forces acting on the stone, its horizontal velocity remains constant throughout its motion. Hence, we can use the equation:
x = v0x * t
Where:
- x is the horizontal displacement
- v0x is the initial horizontal velocity (10.0 m/s)
- t is the time (4 seconds, as calculated in part a)
Plugging in the values, we get:
x = 10.0 * 4
x = 40 meters
Therefore, the stone strikes the ground 40 meters from the base of the cliff.