Answer
(a) $5.62m$
(b) $4.98m$
(c) $5.20m$
Work Step by Step
(a) We can find the required distance as follows:
$t=\frac{x}{vcos\theta}$
$\implies t=\frac{3.05m}{(10.1m/s)cos75^{\circ}}$
$t=1.16s$
Now $y=vsin\theta-\frac{1}{2}gt^2$
We plug in the known values to obtain:
$y=(10.1m/s)sin(75^{\circ})(1.16)-\frac{1}{2}(9.8m/s^2)(1.16s)^2$
$y=4.72m$
$d=\sqrt{x^2+y^2}$
$\implies d=\sqrt{(3.05)^2+(4.72)^2}$
$d=5.62m$
(b) We can find the required distance as
$t=\frac{x}{vcos\theta}$
$t=\frac{4.75}{10.1cos(75^{\circ})}$
$t=1.18s$
Now $d=\sqrt{x^2+y^2}$
$d=\sqrt{(4.75)^2+(1.5)^2}$
$d=4.98m$
(c) We can find the distance just before the projectile lands as
$R=\frac{v^2sin2\theta }{g}$
We plug in the known values to obtain:
$R=\frac{(10.1)^2sin (150^{\circ})}{9.8}$
$R=5.20m$
