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?

Answer 1

From the given data

Horizontal range of the ball, R = 18 m

maximum height above the ground, H = 6 m

maximum height above point of projection, h = (6-1) = 5 m

let Vo is the initial velocity and theta is the angle of projection

we know,

R = vo^2*sin(2*theta)/g

= Vo^2*(2*sin(theta)*cos(theta))/g --(1)

h = vo^2*sin^2(theta)/(2*g) --(2)

take equation(2)/equation(1)

h/R = ( vo^2*sin^2(theta)/(2*g) ) / (Vo^2*(2*sin(theta)*cos(theta))/g )

h/R = tan(theta)/4

5/18 = tan(theta)/4

==> tan(theta) = 1.11

theta = tan^-1(1.11)

= 48 dehrees <<<<<<<<--------------Answer

from equation (1)

vo^2 = 2*g*h/(sin^2(theta))

vo = sqrt(2*g*h)/sin(theta)

= sqrt(2*9.8*5)/sin(48)

= 13.3 m/s <<<<<<<<--------------Answer