Question

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground and experience negligible air resistance. (a) How far from where the child is standing does the ball hit the ground?

Answer 1

Answer:

7.5 m

Explanation:

$v$ = initial speed of the ball = 8 m/s

$\theta$ = angle of launch = 40° deg

Consider the motion along the vertical direction :

$v_{oy}$ = initial velocity along vertical direction = $v Sin\theta$ = 8 Sin40 = 5.14 m/s

$a_{y}$ = acceleration along vertical direction = - 9.8 m/s²

$t$ = time of travel

$y$ = vertical displacement = - 1 m

Using the kinematics equation

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

$- 1 = (5.14)t + (0.5)(- 9.8)t^{2$

$t$ = 1.22 sec

Consider the motion along the horizontal direction :

$v_{ox}$ = initial velocity along horizontal direction = $v Sin\theta$ = 8 Cos40 = 6.13 m/s

$a_{x}$ = acceleration along vertical direction = 0 m/s²

$t$ = time of travel = 1.22 sec

$x$ = horizontal displacement = ?

Using the kinematics equation

$x = v_{ox}t + (0.5)a_{x}t^{2}$

$x = (6.13)(1.22) + (0.5)(0)(1.22)^{2$

$x$ = 7.5 m