A rock is thrown upward with a velocity of 14 meters per second from the top of a 20 meter high cliff, and misses the cliff on the way back down. When will the rock be 12 meters from ground level
x=3.528311 or x=-3.528311
Put it in quadratic equation, so A= -4.9 B=14 C=48,
X= 4.87
To find out when the rock will be 12 meters from ground level, we need to use the kinematic equation for vertical position:
๐ฆ = ๐ฆ0 + ๐ฃ0๐ก + (1/2)๐๐ก^2
Where:
๐ฆ = vertical position of the rock
๐ฆ0 = initial vertical position of the rock
๐ฃ0 = initial vertical velocity of the rock
๐ = acceleration due to gravity (approximately -9.8 m/s^2)
๐ก = time
Given:
๐ฆ0 = 20 meters (since the rock is thrown from a 20 meter high cliff)
๐ฃ0 = 14 m/s (velocity with which the rock is thrown upward)
๐ฆ = 12 meters (the desired height above the ground)
We can rewrite the equation as:
12 = 20 + 14๐ก + (1/2)(-9.8)๐ก^2
Simplifying the equation gives us a quadratic equation:
4.9๐ก^2 + 14๐ก - 8 = 0
To solve this quadratic equation, we can use the quadratic formula:
๐ก = (-๐ ยฑ โ(๐^2 - 4๐๐)) / (2๐)
For this equation, ๐ = 4.9, ๐ = 14, and ๐ = -8. Plugging in the values, we get:
๐ก = (-14 ยฑ โ(14^2 - 4(4.9)(-8))) / (2(4.9))
Simplifying the equation gives us two solutions:
๐ก = (-14 ยฑ โ(196 + 156.8)) / 9.8
๐ก = (-14 ยฑ โ(352.8)) / 9.8
Calculating the square root and simplifying further gives us:
๐ก = (-14 ยฑ 18.8) / 9.8
This gives us two possible times:
๐กโ = (-14 + 18.8) / 9.8 โ 0.48 seconds
๐กโ = (-14 - 18.8) / 9.8 โ -3.47 seconds
Since time cannot be negative in this case, we disregard the negative solution. Therefore, the rock will be at a height of 12 meters from ground level approximately 0.48 seconds after being thrown.
4.96 seconds
just solve the height equation:
20 + 14t - 4.9t^2 = 12