Question

A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground and the fence is 2.0 m high. The ball just clears the fence while still traveling upwards and experiences no significant air resistance. How far is the child from the fence?

Answer 1

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

$x - x_o = u_xt$

$\mathtt{x = u_xt \ \ \ since (x_o = 0)}$  ---- (1)

the equation of the motion y is :

$\mathtt{y - y_o =u_yt - 0.5 gt^2}$

$\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}$

$\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }$

$\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}$

$\mathtt{1 = 5.14t - 4.9t^2}$

$\mathtt{4.9t^2 - 5.14t +1 = 0}$

By using the quadratic formula, we have;

$\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }$

where;

a = 4.9,   b = -5.14     c = 1

$= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }$

$= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }$

$= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }$

$= \mathbf{ 0.791 \ \ OR \ \ 0.258} }$

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

$\mathtt{x = u_x(0.258)}$

$\mathtt{x = ucos 40^0 (0.258)}$

$\mathtt{x = 8 \ cos 40^0 (0.258)}$

$\mathbf{x = 1.581 \ m}$

Thus, the child is 1.581 m far from the fence