A balls is projected horizontally from the top of a hill with a velocity of 20m/s. If it reaches the ground 4 seconds later, what is the height of the hill?
h = 1/2 g t^2 = 1/2 * 9.81 * 16 ... meters
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To find the height of the hill, we can use the equation of motion for vertical motion:
h = vi*t + (1/2)*a*t^2
where:
h = height of the hill
vi = initial vertical velocity (0 m/s, since the ball is projected horizontally)
t = time taken for the ball to reach the ground (4 seconds)
a = acceleration due to gravity (-9.8 m/s^2, assuming no air resistance)
Substituting the given values into the equation:
h = 0 + (1/2)*(-9.8)*(4^2)
h = 0 + (1/2)*(-9.8)*(16)
h = 0 + (-4.9)*(16)
h = -78.4
The height of the hill is -78.4 meters. However, since we typically consider heights as positive values, we can take the absolute value:
h = |-78.4|
h = 78.4
Therefore, the height of the hill is 78.4 meters.
To find the height of the hill, we need to consider the vertical motion of the ball. Since the ball is projected horizontally, its initial vertical velocity is zero and the only force acting on it is the force due to gravity.
We can use the equation for vertical motion:
h = v₀t + (1/2)gt²
where:
h is the height of the hill,
v₀ is the initial vertical velocity (zero in this case),
t is the time of flight (4 seconds in this case),
g is the acceleration due to gravity (approximately 9.8 m/s²).
So, substituting the given values into the equation:
h = 0(4) + (1/2)(9.8)(4)²
h = 0 + (1/2)(9.8)(16)
h = 0 + 4.9(16)
h = 4.9(16)
h = 78.4 meters
Therefore, the height of the hill is 78.4 meters.