Rales l. 150 points) - A foothall is kicked at an angle -37 with a velocity of 2

Question

Rales l. 150 points) - A foothall is kicked at an angle -37 with a velocity of 20 's a. The maximum reached by the bail b. The time of travel before the it hits the groand e How far from away it hits the ground d. The velocity vector at maximum beight e. The acceleration vector at maximam height Assume for simplicity that the ball leaves the foot an ground level 2. In the figure below, a horizontal force F, of magnitude 30.0 N is applied to a 3.00 kg psychology book in an attempt to slide it a distance d an angle -30° with the borizontal. Given that p, 0500 m up a ramp that makes 0.25 und Ak-0.10. During the displacement [30 points] what is the net work done on the book by a. ii. the gravitational force on the book, iii. the frictional force and iv. the normal force on the book? [15 points] If the book has zero kinetic energy at the start of the d what is its speed at the end of the displacement? b. Note:-5 points for free body diagram

Explanation / Answer

Given

football kicked with initial velocity u = 20 m/s at an angle theta = 37 degrees

we can consider it as a projectile  

as the ball reaches the maximum height where the final velocity is zero

and in projectile the horizontal velocity is constant , but the vertical velocity will change at maximum height it become zero

ux = u cos theta

uy = u sin theta

a.

maximum height reached is h_max = u^2 sin^2 theta / 2*g

H_max = 20^2 sin^237/(2*9.8)

H_max = 7.392 m

b. the time of travel before the ball kits the ground is T = 2*u sin theta /g

T = 2*20 sin(37)/9.8 s

T = 2.456 s

c. how far from away it reaches the ground is  

Range R = u^2 sin 2theta /g

R = 20^2 sin (2*37)/9.8 m

R = 39.23 m

d. the velocity vector at maximum height is  

u = ux i + uy j

u = u cos theta i + u sin theta j

u = 20 cos 37 i + 0 j

e. the acceleration vector at maximum height is

there is no acceleration along horizontal as there is no change in horizontal component of velocity so

the vertical component of the velcoity is u sin theta , the acceleration is acceleration due to gravity

a = ax i + ay j

a = 0 i + (-g) j

a = 0 i - g j

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