in a softball game, a batter hits the ball at the velocity of 27m/s and angle of 40 shown below. What is the maximum range of th
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Answer:
The initial velocity in the x-direction with which the branch was thrown is approximately 10.224 m/s
Explanation:
The given parameters of the motion of the branch are;
The height from which the branch is thrown = 3.00 m
The horizontal distance the branch lands from where it was thrown, x = 8.00 m
The direction in which the branch is thrown = Horizontally
Therefore, the initial vertical velocity of the branch, = 0 m/s
The time it takes an object in free fall (zero initial downward vertical velocity) to reach the ground is given as follows;
s = ·t + 1/2·g·t²
Where;
= 0 m/s
s = The initial height of the object = 3.00 m
g = The acceleration due to gravity = 9.8 m/s²
∴ s = 0·t + 1/2·g·t² = 0 × t + 1/2·g·t² = 1/2·g·t²
t = √(2·s/g) = √(2 × 3/9.8) = (√30)/7 ≈ 0.78246
The horizontal distance covered before the branch touches the ground, x = 8.00 m
Therefore, the initial velocity in the horizontal, x-direction with which the branch was thrown, 'uₓ', is given as follows;
uₓ = x/t = 8.00 m/((√30)/7 s)
Using a graphing calculator, we get;
uₓ = 8.00 m/((√30)/7 s) = (28/15)·√30 m/s ≈ 10.224 m/s
The initial velocity in the horizontal, x-direction with which the branch was thrown, uₓ ≈ 10.224 m/s.