Question

A boy standing on the overpass 10 m above the ground throws a ball for his dog. If the boy throws th ball at 303 above the horizontal at 10m/s, a) how long the ball will be in the air, and b) how far from th 1. overpass will it land?

Answer 1

All such projectile motion problems can be solved easily when analysed them as motion in two directions.

We choose the Cartesian Co-ordinate system with the mutually perpendicular x-y axes and we take the initial point of projection as the origin.

Given

Initial velocity $u = 10 m/s$

Angle with horizontal $\theta = 30^{\circ}$

Let us analyse in x and y direction separately.

X direction (Horizontal direction)

Initial velocity in x $u_{x} = ucos\theta = 10*cos(30^{\circ}) = 5\sqrt{3} m/s$

Initial position $x_{i} = 0$

Acceleration in x direction $a_{x} = 0$

That is because acceleration of a body in air is only acceleration due to gravity which is in the downward direction. Here x direction is only horizontal.  

Y direction (Vertical direction)

Initial velocity in y direction $u_{y} = usin\theta = 10*sin(30^{\circ}) = 5m/s$

Initial position $y_{i} = 0$

Acceleration in y direction $a_{y} = -9.8 m/s^2$

This value can also be taken as $a_{y} = -10 m/s^2$ depending on which value the question has given. The most precise value is always 9.8. But taking the g value as 10 is a good close approximation.

Now coming to our analysis.

(a)

The first question wants us to find the time during which the ball is in the air. Since the initial position or the point of projection is above the ground on the overpass at a height of 10m from the ground, we can say that by the time it lands onto the ground the ball will be at the position where its y co-ordinate value will be (-10m)

Final position in y direction $y_{f} = -10m$

This happens when time is equal to the time of duration it was in the air. Let us call it T

So, $t = T$

Now applying the second equation of motion which relates position and time. (Here equations of motion are applicable because the body performs uniformly accelerated motion. That is motion with acceleration constant.

$y_{f} = y_{i} + u_{y}t + \frac{1}{2}a_{y}t^2$

$\Rightarrow -10 = 0 + 5*T + \frac{1}{2}(-10)T^2$

$\Rightarrow 5T^2 -5T -10 = 0$

$\Rightarrow T^2 -T -2 = 0$

This is quadratic equation in T. We can solve for the value of T by factorizing or by using the formula for quadratic equation solutions.

$\Rightarrow (T-2)(T+1) = 0$

$\Rightarrow T = 2 sec$ or $T = -1 sec$
Since time cannot be negative the right solution for the equation is

T = 2 sec.

(b)

The second question needs us to find the distance from the overpass where it lands on the ground.

That is basically nothing but the final x co-ordinate of the point where the ball falls.

When t = T (time duration when it was in air) we can find the final position in x direction using the second equation of motion which relates position and time.

$x_{f} = x_{i} + u_{x}t + \frac{1}{2}a_{x}t^2$

$\Rightarrow x_{f} = 0 + 5\sqrt{3}(2) + \frac{1}{2}(0)(5\sqrt{3})^2$

$\Rightarrow x_{f} = 10\sqrt{3} m$

So it falls at a distance of $10\sqrt{3} m$ from the foot off the overpass. In decimal form it is around 17.32m.

If you solve the same question taking the value of acceleration due to gravity as 9.8, we would slightly different answers. In the place of $a_{y}$ substitute -9.8 instead of the -10 to get the answer if needed that way. Hope you understood. Cheers! :)