Answer all questions below in the exam booklet provided. Not on scrap paper or t

Question

Answer all questions below in the exam booklet provided. Not on scrap paper or the question sheet. Computer, tables or I-phone are not allowed. Be neat. in order to obtain full or partial credit. Print your name on the exam booklet. A projectile is shot from a height of 4m above the ground with initial velocity of 50 m/s directed at the angle of 30 degree with the horizontal. the projectile landed in a pond a distance away. Find the initial x and y components of the velocity (v_x and V_y) Find the position (x, y) of the projectile 2 seconds later What is the maximum height it rises above the ground level? What is the distance from the base (ground) to the point where the ball fell? A long-jumper leaves the ground at an angle of 20.0 degree above the horizontal and at a speed of 11.0 m/s. (a) How far does he jump in the horizontal direction? (Assume his motion is equivalent to that of a particle.) the cliff divers of Acapulco push horizontally from rock platforms about 25 m above the water, but they must clear rocky outcrops at water level that extends out into the water 5.2 m from the base of the cliff directly under the launch pad. find the time taken for a diver to hit the water What should be the initial velocity of a diver?

Explanation / Answer

1)

(a)

Vx = v*costheta = 50*cos30 = 43.3 m/s

vy = v*sintheta = 50*sin30 = 25 m/s

(b)

x = vx*t + (0.5*ax*t^2)

x = 43.3*2 + 0 = 86.6 m

y = vy*t + (0.5*ay*t^2)

y = (25*2) - (0.5*9.8*2^2) = 30.4 m

(x , y ) = (86.6 , 30.4)

(c)

at maximm final velocity vfy = 0

vfy^2 - vy^2 = 2*ay*h

0 - 25^2 = -2*9.8*h

h = 31.9 m

(d)

after the ball reaches the ground

along vertical displacement dy = 0

dy = vy*t + (0.5*ay*T^2)

0 = 25*T - (0.5*9.8*T^2)

T = 5.12 s

along horizontal

x = vx*T = 43.3*5.12 = 221.7 m <<<<<<-----answer

================

(2)

(a)

after the ball reaches the ground

along vertical displacement dy = 0

dy = vy*t + (0.5*ay*T^2)

0 = (11*sin20*T) - (0.5*9.8*T^2)

T = 0.77 s

along horizontal

x = vx*T = 11*cos20*0.77 = 7.96 m <<<<<<-----answer

===========

3)

along vertical

dy = voy*t + (1/2)*ay*T^2

-25 = 0 - (0.5*9.8*T^2)

Time T = 2.26 s <<<<====answer

(b)

along horizontal

x = vx*T

5.2 = vx*2.26

vx = 2.3 m/s <<<<<=====answer

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