Question

You're 6.0 m from one wall of a house. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground.(Figure 1) Assume the overhang of the roof is negligible, so that you may assume the edge of the roof is 6.0 m from you and 6.0 m from your friend.

What minimum speed will allow the ball to clear the roof?

At what angle should you toss the ball?

Please explain for 5stars

Answer 1

Concept and reason

The main concept used is equation of projectile motion.

Initially, use the trigonometry to calculate the maximum height from the graph, Later, use the expression of the velocity of the projectile. Finally use the expression of the maximum height of the projectile and range to calculate the angle.

Fundamentals

Consider an object projected with an initial velocity u at angle with horizontal direction. During the time of flight the horizontal displacement is R and the maximum height attained by the projectile is denoted by H.

The angle of projection of the projectile is given by following formula:

The minimum speed u with which projectile should be projected so that it clears off the roof is given by following expression:

Here g denotes the acceleration due to gravity.

[Starting hint]

Calculate the angle of projection by equating the formula for horizontal range and maximum height attained by the projectile.

The horizontal range of an object projected at some angle is given by following expression:

In addition, the maximum height H attained by the projectile is given by following expression:

Divide and solve the expression given above for angle of projection as follows:

So, the angle of projection is given as follows:

The horizontal range of the projectile is given as the sum of the distance a of the point of projection from the nearest wall, distance b of the point of landing from the nearest wall and the width w of the house and given as follows:

$R=a+b+w $

Substitute for each a, b and w in the last expression as follows:

Let one edge of the overhang of the roof makes angle with horizontal and y be the height of the peak of the roof from level AB.

The height y is given below:

Substitute for w and for as follows:

The maximum height attained by the projectile is given as follows:

$H=h+h+y $

Here, is the height of the point of projection from ground and height is shown in the figure.

Substitute for , for and for as below:

The angle of projection is given by following expression:

Substitute for R and for H as follows:

Minimum speed of the ball for crossing cross the wall is given as follows:

Substitute for R, for g and for angle of projection as follows:

(b)

The horizontal range of an object projected at some angle is given by following expression:

In addition, the maximum height H attained by the projectile is given by following expression:

Divide and solve the expression given above for angle of projection as follows:

So, the angle of projection is given as follows:

The horizontal range of the projectile is given as the sum of the distance a of the point of projection from the nearest wall, distance b of the point of landing from the nearest wall and the width w of the house and given as follows:

Substitute for each a, b and w in the last expression as follows:

Let one edge of the overhang of the roof makes angle with horizontal and y be the height of the peak of the roof from level AB.

The height y is given below:

Substitute for w and for as follows:

The maximum height attained by the projectile is given as follows:

Here, is the height of the point of projection from ground and height is shown in the figure.

Substitute for , for and for as below:

The angle of projection is given by following expression:

Substitute for R and for H as follows:

Ans: Part a

The minimum speed of the ball to clear the roof is .

Part b

For the ball to cross the roof, the angle of projection is .