Question

5) A snowball is thrown at an angle of 39° with an initial speed of 20 m/s and an initial height of 3 m. Determine the maximum height reached by the snowball. A) 14.63 m B) 10.91 m C) 10.39 m D) 11.08 m

Answer 1

$\text{Given is :-} \newline \text{Initial velocity }=20m/s \newline \text{Angle =}39^\circ \newline \text{thus the horizontal and vertical components of the initial velocity are} \newline u_x = u cos \theta \text{ and } u_y = u sin \theta \newline \text{by plugging the values we get} \newline u_x = 20cos (39 ^\circ) \text{ and } u_y = 20sin (39 ^\circ) \newline \text{which gives us} \newline u_x = 15.5429 m/s \text{ and } u_y = 12.5864 m/s$

Now,

$\text{Using Third equation of motion} \newline \newline v^2 = u^2 + 2as \newline \newline \text{by plugging all the values we get} \newline \newline (0 \hspace{1mm} m/s)^2=(12.5864 \hspace{1mm} m/s)^2+2(-9.8 \hspace{1mm} m/s^2)(s \hspace{1mm} m) \newline \newline \text{Which gives us} \newline \newline s = 8.083 m$

Thus the total height reached by the ball is

$h = \text{ initial height} + s$

or

which gives us

$\boxed{h = 11.083m }$

Hence,

option - D is correct answer.