Answer
$a). $(1) longer time,
$b). $final velocity=(11.3x-7.5y)m/s
Range=13.3m
Work Step by Step
a). The time taken will be longer, because the shot is taken from 2 m above ground level. it will take more time to traverse the additional distance due to height.
b). $v_{y}^{2}=(v_{0}sin\theta)^{2}+2gh$
$v_{y}=7.5m/s$
Now, for part (a).,
$t$ when launched from ground = $\frac{2\times 12\times sin20^{\circ}}{9.8}=0.84s$
$t$ when launched from height = $\frac{7.5+12\times sin20^{\circ}}{9.8}=1.18s$
Hence, it shows the time taken by shot taken from height is greater.
Now, $v_{x}=vcos\theta=12cos20^{\circ}=11.3m/s$
So, final velocity =$(11.3x-7.5y)m/s$
Range=$vcos\theta \times t=11.3\times 1.18=13.3m$
