A person who is 1.2 m tall throws a ball with a speed of 28 m/s at an angle of 7

Question

A person who is 1.2 m tall throws a ball with a speed of 28 m/s at an angle of 78 degrees above the horizontal, as shown in the diagram below.
h = 1.2 m, v = 28 m/s, ? = 78 degrees

Use the acceleration due to gravity as g = 9.81 m/s2.

1)What are the x and y components of the initial velocity?

2)What are the x and y components of the ball's velocity when it is at its maximum height?

3)What are the x and y components of the ball's velocity the instant before it hits the ground?

4)What is the ball's speed (the magnitude of the velocity) the instant before it hits the ground?

5)What is the time of flight; that is, the time taken for the ball to hit the ground after it has left the person's hand?

Explanation / Answer

1)

x comp = vcos 78 = 28 cos 78 = 5.82 m/s

y comp = vsin 78 = 28 sin 78 = 27.39 m/s

2)

at maximum height the vertical comp will be zero

so y-comp = 0

horizontal x-comp will remain same

x comp = 5.82 m/s

3)

x component stays the same

x-comp = 5.82 m/s

y-comp

v = sqrt( u^2+2ah)

u = -27.39 m/s

a = 9.81 m/s^2

h= 1.2 m

v= sqrt (-27.39^2 +2*9.81*1.2)

v= 27.81 m/s

4)

us conservation of energy

initial

0.5*m*28*28 + m*9.81*1.2

final = 0.5*m*v*v

equate

0.5*v*v = 0.5*28*28 + 9.81*1.2

v= 28.417 m/s

5)

vertical

u = -27.39m/s

v=27.81 m/s

a=9.81m/s^2

v=u+at

t=27.81-(-27.39) / 9.81

t=5.63 sec

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