Answer
$1.75N\cdot s$
Work Step by Step
The vertical and horizontal components of velocity are: $v_x=vcos60^{\circ}$
and $v_y=v sin60^{\circ}$
For vertical distance, we use the equation of motion
$y=vsin60^{\circ}\times t-\frac{1}{2}gt^2$
but at the highest point $y=0$
$\implies 0=vsin60^{\circ}\times -\frac{1}{2}gt^2$
$\implies t=\frac{2vsin60^{\circ}}{g}$.....eq(1)
Now $v_x=vcos60^{\circ}\times t$
$\implies 12=vcos60^{\circ}\times t$
$\implies t=\frac{12}{vcso60^{\circ}}$......eq(2)
Comparing eq(1) and eq(2), we obtain:
$\frac{12}{vcos60^{\circ}}=\frac{2vsin60^{\circ}}{g}$
After simplifying the above equation, we have
$v=11.66m/s$
Now the impulse can be determined as:
$m(v_f-v_i)=I$
We plug in the known values to obtain:
$\implies I=0.15\times 11.66$
$I=1.75N\cdot s$
