Respuesta :
Answer:
The maximum height of the ball is 2 m.
Explanation:
Given that,
Mass of ball = 50 g
Height = 1.0 m
Angle = 30°
The equation is
[tex]y=\dfrac{1}{4}x^2[/tex]
We need to calculate the velocity
Using conservation of energy
[tex]\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}[/tex]
Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero
[tex]\Delta U_{i}=\Delta K_{f}[/tex]
Put the value into the formula
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
Put the value into the formula
[tex]50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2[/tex]
[tex]v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}[/tex]
[tex]v=\sqrt{19.6}[/tex]
[tex]v=4.42\ m/s[/tex]
We need to calculate the maximum height of the ball
Using again conservation of energy
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
Here, h = y highest point
Put the value into the formula
[tex]\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h[/tex]
[tex]y=\dfrac{0.5\times(4.42)^2}{9.8}[/tex]
[tex]y=0.996\ m[/tex]
Put the value of y in the given equation
[tex]y=\dfrac{1}{4}x^2[/tex]
[tex]x^2=4\times0.996[/tex]
[tex]x=\sqrt{4\times0.996}[/tex]
[tex]x=1.99\ m\ \approx 2 m[/tex]
Hence, The maximum height of the ball is 2 m.
