Question

A ball is launched with an initial velocity of 3m/s, at an angle of 40 degrees above the horizontal, and from a height of 0.5m. How far will it travel horizontally before it hits the ground?

Answer 1

Answer:

The ball travels 1.31 m horizontally before hitting the ground.

Explanation:

Given:

Initial velocity of the ball is, $u=3\ m/s$

Angle of projection is, $\theta=40\ degree$

Initial height of the ball is, $y_0=0.5\ m$

Final height of the ball is, $y=0\ m$

Now, horizontal  and vertical components of initial velocity are given as:

$u_x=u\cos\theta\\\\u_y=u\sin\theta$

Plug in the given values and find $u_x\ and\ u_y$. This gives,

$u_x=3\ m/s\times \cos(40)\\\\u_x=2.3\ m/s\\\\u_y=3\ m/s\times \sin(40)\\\\u_y=1.9\ m/s$

Now, there is acceleration due to gravity acting in the vertical direction. The direction is downward. So, $g=-9.8\ m/s^2$.

Now, using equation of motion in the vertical direction, we have:

$y-y_0=u_yt+\frac{1}{2}a_y t^2$

Plug in the given values and solve for time 't'. This gives,

$0-0.5=1.9t-0.5\times 9.8\times t^2\\\\-0.5=1.9t-4.9t^2\\\\4.9t^2-1.9t-0.5=0$

On solving the above quadratic equation, we get:

$t=0.57\ s\ or\ t=-0.18\ s$

Ignoring the negative result as time can't be negative. So, time taken by the ball to reach the ground is 0.57 s.

Now, as the ball moves, there is no force acting on it in the horizontal direction. So, the acceleration in the horizontal direction is 0.

Now, we know that, when acceleration is zero, the body moves with a constant speed.

Hence, distance traveled in the horizontal direction is given as:

Horizontal distance = Horizontal component of initial velocity × Time

$R=u_x\times t$

Plug in the given values and solve for 'R'. This gives,

$R=2.3\times 0.57\\\\R=1.31\ m$

So, the ball travels 1.31 m horizontally before hitting the ground.