A rock is thrown downward from an unknown height above the ground with an initial speed of 21 m/s. It strikes the ground 8 seconds later. Determine the initial height of the rock above the ground. The acceleration of gravity is 9 m/sĀ².
Answer in units of meters.
Answer (3)
Given: initial velocity (u) = 21 m/s time (t) = 8 seconds acceleration (a) = 9 m/sĀ² distance (s) = ???????? Now, you can use the formula, s = u t + 2 1 ā a t 2
Now, plug the values in the formula, and you get:
s = 21 ā 8 + 2 1 ā ā 9 ā 8 2
s = 21 ā 8 + 2 1 ā ā 9 ā 64
s = 21 ā 8 + 2 1 ā ā 576
s = 21 ā 8 + 288
s = 168 + 288
s = 456 m e t ers
SO THE INITIAL HEIGHT OF THE ROCK WAS 456 METERS.
To find the initial height of the rock above the ground, we can use the kinematic equation for vertical motion:
h = h0 + v0t + (1/2)gt2
Where: h = final height (0 m) h0 = initial height v0 = initial velocity (21 m/s) t = time (8 s) g = acceleration due to gravity (-9.8 m/s2)
Plugging in the given values:
0 = h0 + (21 m/s)(8 s) - (1/2)(9.8 m/s2)(8 s)2
Simplifying the equation:
0 = h0 + 168 m - 313.6 m
h0 = -168 m + 313.6 m
h0 = 145.6 m
Therefore, the initial height of the rock above the ground is 145.6 m.
The initial height of the rock above the ground is 145.6 meters, calculated using the kinematic equation for vertical motion. By accounting for the initial speed and the distance fallen due to gravity over 8 seconds, we found that the final height at ground level is zero. Therefore, the calculation results in an initial height of 145.6 meters.
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