Answer
(a) The highest point above the ground $= 2.8 m$
(b) horizontal displacement $= 2 m$
Work Step by Step
(a) Here we use equation 3.6b $V^{2}=u^{2}+2aS$ to find the highest vertical displacement of the skateboard.
$\uparrow V^{2}=u^{2}+2aS$ ; Let's plug known values into this equation.
$0^{2}=(6.6\space m/s\times sin58^{\circ})^{2}+2\times (-9.8\space m/s^{2})S$
$S=\frac{(5.6\space m/s)^{2}}{19.6\space m/s^{2}}=1.6\space m$
The, highest point reached by skateboard = 1.2 m + 1.6 m = 2.8 m
(b) Here we use the equation $V=u+at$ to find the time taken to reach its maximum height.
$\uparrow V=u+at$ ; Let's plug known values into this equation.
$0=6.6\space m/s\times sin58^{\circ}+(-9.8\space m/s^{2})t$
$t=0.57\space s$
Let's apply the equation $S=ut$ to find the skateboarder's horizontal displacement.
$\rightarrow S=ut$ ; Let's plug known values into this equation.
$R=6.6\space m/s\times cos58^{\circ}\times 0.57\space s=2m$
