Respuesta :
Answer:
14.14 m/s
Explanation:
mass of ball, m 0.150 kg
initial velocity, u = 20 m/s
g = 10 m/s²
The final velocity at the height is zero. Total height is y2.
Use third equation of motion,
v² = u² - 2 gh
0 = 20 x 20 - 2 x 10 x y2
y2 = 20 m
Let v is the velocity at height y2/2 .
v² = u² - 2 gh
v² = 20 x 20 - 2 x 10 x 10
v² = 400 - 200
v = 14.14 m/s
Thus, the velocity at half of the height is 14.14 m/s.
The speed of the ball will be 14.14 m/s.Speed is represented as the pace of change of the distance or the height attained.
What is speed ?
Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its si unit is m/sec.
The given data in the problem is;
m is the mass of all= 0.150 kg
u is the initial velocity= 20 m/s
g is the gravitational acceleration = 10 m/s²
v is the speed of ball it is at a height of y2/2 =?
According to the Newtons third equation of motion for the final velcity v=0 m/sec;
[tex]\rm v^2 = u^2-2gh \\\\\rm 0^2 = (20)^2-2\times 9.81 \times y_2 \\\\ \rm y_2 = 20\ m[/tex]
If v is the velocity for the height [tex]\rm \frac{y_2 }{2}[/tex]
[tex]\rm v^2 = u^2-2gh \\\\\rm v^2 = (20)^2-2\times 9.81 \times 10 \\\\ \rm v = 14.14 \ m/sec[/tex]
Hence the velocity at half of the height is 14.14 m/s.
To learn more about the sped refer to the link;
https://brainly.com/question/7359669
