Answer:
a) The height of the power line is 4.7 m
b) The initial velocity is 9.8 m/s
Step-by-step explanation:
The equation for the height of the ball is the following:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height of the ball
y0 = initial height
v0 = initial velocity
t = time
g = acceleration due to gravity
Let´s place the origin of the frame of reference at the throwing point so that y0 = 0
b) We know that when the time is 0.80 s and 1.2 s the height is the same. Then:
y = y0 + v0 · t + 1/2 · g · t²
y = v0 · 0.80 s - 1/2 · 9.8 m/s² · (0.80 s)²
y = v0 · 1.2 s - 1/2 · 9.8 m/s² · (1.2 s)²
v0 · 0.80 s - 1/2 · 9.8 m/s² · (0.80 s)² = v0 · 1.2 s - 1/2 · 9.8 m/s² · (1.2 s)²
v0 · 0.80 s - 3.136 m = v0 · 1.2 s - 7.056 m
v0 · 0.80 s - v0 · 1.2 s = - 7.056 m + 3.136 m
-0.4 s · v0 = -3.92 m
v0 = -3.92 m / -0.4 s
v0 = 9.8 m/s
The initial velocity is 9.8 m/s
a) Now, we can calculate the height of the power line:
y = y0 + v0 · t + 1/2 · g · t²
y = 9.8 m/s · 1.2 s - 1/2 · 9.8 m/s² · (1.2 s)²
y = 4.7 m
The height of the power line is 4.7 m