1/A person standing 20.0 meters away from a hill throws a stone from the ground

Question

1/A person standing 20.0 meters away from a hill throws a stone from the ground level at an angle of 60.0°above the horizontal with an initial speed of 25.0 m/s. The time it takes the stone to reach the hill and the height at which the stone strikes the hill are:

t = 1.4 s, y = 18.6 m

t = 1.6 s, y = 22.1 m

t=2.6s, y=11.5m

t=6.8s, y=34.2m

2/Which of the following statements is correct?

Mass is a vector quantity

Velocity is a scalar quantity

Time is a scalar quantity

Acceleration is neither a vector nor a scalar quantity.

A ball is launched from ground level at 30 m/s at an angle of 35 above the horizontal.
   (a) Find the speed of the ball when it reaches the maximum height.
   (b) How far does it go (horizontally) before it is at ground level again?

(a) 24.6 m/s,    (b) 86.3 m

(a) 15 m/s ,    (b) 43.2 m

(a) 17.5 m/s, (b) 90.8 m

(a) 0,    (b) 86.3 m

t = 1.4 s, y = 18.6 m

t = 1.6 s, y = 22.1 m

t=2.6s, y=11.5m

t=6.8s, y=34.2m

2/Which of the following statements is correct?

Mass is a vector quantity

Velocity is a scalar quantity

Time is a scalar quantity

Acceleration is neither a vector nor a scalar quantity.

A ball is launched from ground level at 30 m/s at an angle of 35 above the horizontal.
   (a) Find the speed of the ball when it reaches the maximum height.
   (b) How far does it go (horizontally) before it is at ground level again?

(a) 24.6 m/s,    (b) 86.3 m

(a) 15 m/s ,    (b) 43.2 m

(a) 17.5 m/s, (b) 90.8 m

(a) 0,    (b) 86.3 m

Explanation / Answer

only one question can be posted as per chegg guidlines

1,

Answer is t = 1.6 s, y = 22.1 m

Initial Horizontal and vertical components of velocities are

Vox=25Cos60

Voy=25Sin60

Horizontal distance travelled

X=Voxt

=>t=20/25Cos60 =1.6 s

From

Y=Yo+Voyt-(1/2)gt2

Y=0+(25sin60)(1.6)-(1/2)(9.8)(1.6)2

Y=22.1 m

2.

Answer is Time is a scalar quantity

Velocity is vector quantity and mass is scalar quantity

3.

ANswer is (a) 24.6 m/s,    (b) 86.3 m

Vox=30Cos35

Voy=30Sin35

a)

at maximum height

Vfy=0

and Vfx=Vox =30Cos35=24.6 m/s

speed of ball at maximum height

V=sqrt[24.62+0]=24.6 m/s

b)

From

Y=Yo+Voyt-(1/2)gt2

when ball hits ground Y=0

0=0+(30SIn35)t-(1/2)(9.8)t2

t=3.51 s

Horizontal distance travelled

X=Voxt =(30cos35)(1.874) =86.3 m

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