Answer:
Sorry this isn’t going to be any help. You don’t have any statement that I’m able to see.
Explanation:
We are given the equation:
<span>x = 11t^2
</span>
We use that equation to calculate for the distance traveled.
For (a)
At t=2.20 sec,
x =53.24 meters
At t=2.95 sec,
x =95.73 meters
Velocity = (95.73 meters - 53.24<span> meters) / (2.95 s - 2.20 s ) = 56.65 m/s
</span>For (b)
At t=2.20 sec,
x =53.24 meters
At t=2.40 sec,
x =63.36 meters
Velocity = (63.36 meters - 53.24<span> meters) / (2.40 s - 2.20 s ) = 50.6 m/s</span>
A=m/s^2(meter per second square)
Work=joule
Given that,
Height =1.5 m
Angle = 45°
We need to find the greater speed of the ball
Using conservation of energy
Here, initial velocity and final potential energy is zero.
Put the value into the formula
Hence, the greater speed of the ball is 5.42 m/s.
