Answer to Question #255990 in Mechanics | Relativity for drei

Question

A ball is thrown horizontally from a building  high with a speed . Find the (a) vertical and horizontal components of the ball’s intial velocity, (b) time of flight, (c) distance from the foot of the building where the ball will strike the ground, and (d) velocity when the ball will strike the ground.

Answer 1

Explanations & calculations

a)

• To calculate the components of the velocity, you need to deal with the angle at which the ball is thrown. For this case, the ball is thrown horizontally, which means the angle it creates with the horizontal is zero.
• Therefore, all the velocity (initial) is directed horizontally leaving no vertical speed vertically so that the horizontal component is the all: $\small v$ (not provided but I assume a mistake has occurred) and the vertical component is zero.

b)

• Since the projectile occurs under the constant acceleration: g (9.81m/s^2), any of the 4 motion equations can be used appropriately.
• For the downward vertical movement of the ball's projectile, you can apply $\small s = ut+1/2at^2$ as you are given the height of the building, zero initial vertical speed and the gravitational acceleration: g.
• Time of flight is how long it takes for the ball to fall on the ground from the building top. During this time it travels both horizontally and vertically.

$\qquad\qquad \begin{aligned} \small \downarrow\\ \small h&=\small 0t+0.5\times g\times t^2\\ \small t&=\small \sqrt{\frac{2h}{g}} \end{aligned}$

c)

• As said above, it travels some distance horizontally during the time of flight and it is called the range (R) of the flight.
• To calculate that, you can apply the same equation used for the previous part yet, it simplifies down to $\small s = ut$ as there is no acceleration the ball experiences horizontally.

$\qquad\qquad \begin{aligned} \to\\ \small R&=\small v\times t\\ &=\small v\sqrt{\frac{2h}{g}} \end{aligned}$

d)

• When the ball strikes the ground, it has a velocity directed at some angle to the ground thus having both verticle and horizontal components.
• To calculate the vertical component (probably directed vertically downward), you can use $\small v = u+at$.
• To calculate the horizontal velocity component you do not need to sweat. It is the same as the initial, $\small v_x=v$ as there is no acceleration horizontally to increase the ball's velocity during the time of flight.
• Once you have calculated both, the final velocity can be found using the Pythagoras theorem as the two components are normal to each other.

$\qquad\qquad \begin{aligned} \downarrow\\ \small v_y&=\small 0+gt\\ &=\small g\sqrt{\frac{2h}{g}}\\ &=\small \sqrt{2gh}\\ \\ \small v_f&=\small \sqrt{v_x^2+v_y^2}\\ &=\small \sqrt{v^2+2gh} \end{aligned}$

• Since you have been asked for the velocity, not just the speed, you need to mention the direction at which the final speed vector is directed. To calculate that, the tangent of the vector triangle that is formed by $\small v_x, v_y\,\&\,v_f$ can be taken into consideration (they align in a right triangle).

$\qquad\qquad \begin{aligned} \small \tan\theta&=\small \frac{v_y}{v_x}\\ \small \theta&=\small \tan^{-1}\Big(\frac{\sqrt{2gh}}{v}\Big) \end{aligned}$

• These are the universal basics that are common to all of the projectile motions, yet you need to know the initial horizontal speed and height of the building to work this problem.