Answer:
0.20387 meters or 20.4 cm
Explanation:
This is a kinetic energy to potentially energy question
write it as follows
m = 0.2kg
v0 = 2m/s
h= ?
g = (whatever value your class uses typically 9.81 or 10)
KE = 1/2 * m*v^2
= 1/2 * 0.2*2^2
= 0.5*0.2*4
= 0.4 joules
PE = mgh
=0.2*9.81*h
since there is a full theoretical 100% energy conversion at the peak of the balls moment PE=KE
0.2*9.81*h = 0.4
h = 0.4/(0.2*9.81)
h= 0.20387 meters
20.4 cm
Answer:
[tex] \textsf{0.20387 m or 20.387 cm} [/tex]
Explanation:
To find the maximum height reached by the ball, we can use the following kinematic equation for vertical motion under gravity:
[tex] \Large\boxed{\boxed{h = \dfrac{{u^2 \sin^2 \theta}}{{2g}} }}[/tex]
Where:
Given:
Substituting the given values into the formula:
[tex] h = \dfrac{{(2 \, \textsf{m/s})^2 \sin^2 (90^\circ)}}{{2 \times 9.81 \, \textsf{m/s}^2}} [/tex]
Since [tex] \sin(90^\circ) = 1 [/tex], we have:
[tex] h = \dfrac{{(2 \, \textsf{m/s})^2 \times 1}}{{2 \times 9.81 \, \textsf{m/s}^2}} [/tex]
[tex] h = \dfrac{{4 \, \textsf{m}^2/\textsf{s}^2}}{{19.62 \, \textsf{m/s}^2}} [/tex]
[tex] h = \dfrac{{0.204 \, \textsf{m}^2/\textsf{s}^2}}{{19.62 \, \textsf{m/s}^2}} [/tex]
[tex] h \approx 0.2038735983690 \, \textsf{m} [/tex]
[tex] h \approx 0.20387 \textsf{ m (in 5 d.p. or } 20.387 \textsf{ cm} [/tex]
So, the maximum height reached by the ball is approximately [tex] \textsf{0.20387 m or 20.387 cm} [/tex].
