Answer
(a) 9.09 m (b) 1.06 s
Work Step by Step
Let's apply equation 3.3b $V=u+at$ in the vertical direction to find the time taken to reach its maximum height.
$\uparrow V=u+at$ ; Let's plug known values into this equation.
$0=10\space m/s\times sin31^{\circ}+(-9.8\space m/s^{2})t$
$t=0.53\space s$
Let's apply the equation $S=ut$ in the horizontal direction to find the range of the leap
$\rightarrow S=ut$ ; Let's plug known values into this equation.
$R=10\space m/s\times cos31^{\circ}\times0.53\space s\times2=9.09\space m$
(b) We know that the time for the projectile to rise from ground level to its maximum height is equal to the time for the projectile to fall back to ground level. So the greyhound is in the air for 2(0.53 s) = 1.06 s
