Respuesta :
Answer:
v = 8.96 m/s
Explanation:
Initial speed of the ball, u = 10 m/s
It caught 1 meter above its initial position.
Acceleration due to gravity, [tex]g=-9.8\ m/s^2[/tex]
We need to find the final speed of the ball when it is caught. Let is equal to v. To find the value of v, use third equation of motion as :
[tex]v^2-u^2=2as[/tex]
[tex]v^2=2as+u^2[/tex]
[tex]v^2=2(-9.8)\times 1+(10)^2[/tex]
v = 8.96 m/s
So, the speed of the ball when it is caught is 8.96 m/s. Hence, this is the required solution.
Answer:
8.96 m/s, upward direction
Explanation:
Given that, the initial velocity of the ball is,
[tex]u=10m/s[/tex]
And the acceleration in the downward direction is positive but in this situation the acceleration will be negative so,
[tex]a=9.8\frac{m}{s^{2} }[/tex]
And according to question vertical displacement is,
[tex]s=1m[/tex]
Now suppose v be the final velocity of the ball.
Applying third equation of motion,
[tex]v^{2}=u^{2}+2as[/tex]
Here, u is the initial velocity, a is the acceleration, s is the displacement.
Substitute all the variables.
[tex]v=\sqrt{10^{2}+2(-9.8)\times 1 } \\v=\sqrt{80.4}\\ v=8.96\frac{m}{s}[/tex]
Therefore, the speed of ball when it is caught is 8.96 m/s in the upward direction.
