Question

A man stands on the roof of a building of height 14.0 mm and throws a rock with a velocity of magnitude 29.0 m/sm/s at an angle of 29.0 ∘∘ above the horizontal. You can ignore air resistance.Calculate (a) the maximum height above the roof that the rock reaches;(b) the speed of the rock just before it strikes the ground; and(c) the horizontal range from the base of the building to the point where the rock strikes the ground.(d) Draw x-t, y-t, vx−t, and vy−t graphs for the motion.

Answer 1

a) y_f = 10.076 m

b) V_y,f = 21.734 m/s

c) s_x,f = 72.705 m

Step-by-step explanation:

Given:

- Initial height from which rock is thrown y(0) = 0

- Initial velocity V_x,i = 29*cos(29) = 25.364 m/s

- Initial velocity V_y,i = 29*sin(29) = 14.06 m/s

Find:

(a) the maximum height above the roof that the rock reaches

(b) the speed of the rock just before it strikes the ground

(c) the horizontal range from the base of the building to the point where the rock strikes the ground.

Solution:

a)

The maximum height above the roof can be calculated when V_y,f = 0.

For which we can use the third equation of motion as follows:

V_y,f^2 - V_y,i^2 = -2*g*y_f

- Input values:

0 - 14.06^2 = -2*(9.81)*y_f

y_f = 14.06^2 / 2*(9.81)

y_f = 10.076 m

b)

The speed of the ball when it hits the ground can be calculated when y_f = -14 m . Again using the third equation of motion we have:

V_y,f^2 - V_y,i^2 = -2*g*y_f

- Input Values:

V_y,f^2 = -2*g*y_f + V_y,i^2

V_y,f^2 = -2*(9.81)*(-14) + 14.06^2

V_y,f = sqrt( 472.3636)

V_y,f = 21.734 m/s

c)

For this part we will compute the time taken by the rock to come back to the initial position. Using second equation of motion we have:

s_y,f - s_y,i = V_y,i*t -4.905*t^2

- Plug in values:

0 - 0 = 14.06*t -4.905t^2

t = 14.06 / 4.905 = 2.87 s

- Now we will compute the distance traveled in x - direction for the above calculated time t = 2.87 s. Use second equation of motion for x direction:

s_x,f = s_x,i + V_x,i*t

- Plug in values:

s_x,f = 0 + 2.87*25.364

s_x,f = 72.705 m