Answer
$(a)\space 85\space m$.
$(b)\space 610\space m$
Work Step by Step
Let's find the flight time of the ball on earth. Here we use the equation $S=ut+\frac{1}{2}at^{2}$ in the vertical direction.
$\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$0=45\space m/s\space \times sin29^{\circ}t+\frac{1}{2}(-9.8\space m/s^{2})t^{2}$
$t=\frac{2\times(21.82\space m/s)}{9.8\space m/s^{2}}=4.45\space s$
The flight time on the distant planet $=3.5\times4.45\space s=15.6\space s$
We know that this is twice the amount of time required for the ball to reach its maximum height, which is 7.80 s
(a) Let's apply equation $S=\frac{(u+V)}{2}t$ in the vertical direction to the ball on the distant planet to find its maximum height.
$\uparrow S=\frac{(u+V)}{2}t$ ; Let's plug known values into this equation.
$H=\frac{45\space m/s\times sin29^{\circ}}{2}\times 7.8\space s=85\space m$
Maximum height = 85 m
(b) Let's apply equation $S=ut$ in the horizontal direction to find the range of the ball on the distant planet.
$\rightarrow S=ut$
$R=45\space m/s \times cos29^{\circ}\times 15.6\space s=610\space m$
Range = 610 m