Question

A student standing on a knoll throws a snowball horizontally 4.5 meters above the level ground toward a smokestack 15 meters away. The snowball hits the smokestack 0.65 second after being released. [Neglect air resistance.]
Approximately how far above the level ground does the snowball hit the smokestack?

Answer 1

Answer:

2.4 m

Explanation:

Consider the motion along the vertical direction

$y_{o}$ = initial position of ball above the ground = 4.5 m

$t$ = time taken by the ball to hit the smokestack = 0.65 s

$v_{oy}$ = initial velocity of the ball along vertical direction

$a_{y}$ = acceleration due to gravity = - 9.8 m/s²

$y$ = position of ball at the time of hitting the smokestack

Using the kinematics equation

$y = y_{o} + v_{oy} t + (0.5) a_{y} t^{2}$

inserting the above values

$y = 4.5 + (0) (0.65) + (0.5) (- 9.8) (0.65)^{2} \\y = 2.4 m$