Answer:
0.375 m
Explanation:
To find the maximum height of the deer's jump, we can use the following kinematic equation for projectile motion:
[tex]\sf h = \dfrac{{v_0^2 \sin^2(\theta)}}{{2g}} [/tex]
where:
Given:
[tex]\sf v_0 = 3.75 \, \textsf{m/s} [/tex] ,
[tex]\sf \theta = 46.3^\circ [/tex],
[tex]\sf g = 9.8 \, \textsf{m/s}^2[/tex] ,
Substitute these values into the equation:
[tex]\sf h = \dfrac{{(3.75 \, \textsf{m/s})^2 \sin^2(46.3^\circ)}}{{2 \times 9.8 \, \textsf{m/s}^2}} [/tex]
Calculate:
[tex]\sf h = \dfrac{{ 14.0625 \, \textsf{m}^2/s^2 \times \sin^2(46.3^\circ)}}{{19.6 \, \textsf{m/s}^2}} [/tex]
[tex]\sf h = \dfrac{{14.0625 \, \textsf{m}^2/s^2 \times 0.5226814941 }}{{19.6 \, \textsf{m/s}^2}} [/tex]
[tex]\sf h \approx \dfrac{{7.351515214 \, \textsf{m}^2/s^2}}{{19.6 \, \textsf{m/s}^2}} [/tex]
[tex]\sf h \approx 0.3750773068 \, \textsf{m} [/tex]
[tex]\sf h \approx 0.375 \, \textsf{m ( in 3 d.p.)} [/tex]
So, the maximum height of the deer's jump is approximately 0.375 m
