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PHYSICS, M1 EQ-Bank 6 MC

A skier launches off a small ramp that is 2.0 m above the ground, moving horizontally. She lands 16 m away from the base of the ramp, and it takes her 0.8 seconds to reach the ground.

Assume air resistance is negligible and the horizontal velocity remains constant throughout the jump.

What was her speed at take-off, rounded to the nearest integer?

  1. 2.5 m/s
  2. 12.8 m/s
  3. 15 m/s
  4. 20 m/s
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\(D\)

Show Worked Solution
  • The increasing vertical velocity of the skier is not required when calculating the skier’s initial horizontal velocity.
  • The horizontal velocity of the skier can be calculated:

\(v_h = \dfrac{d_h}{t} = \dfrac{16}{0.8} = 20\ \text{m/s}\)

\(\Rightarrow D\)

PHYSICS, M1 EQ-Bank 5 MC

A student throws a ball with a speed of 20 m/s at an angle of 40° above the horizontal. What are the horizontal and vertical components of the ball’s velocity?

  1. Horizontal: 15.3 m/s, Vertical: 12.9 m/s
  2. Horizontal: 12.9 m/s, Vertical: 15.3 m/s
  3. Horizontal: 20.0 m/s, Vertical: 0.0 m/s
  4. Horizontal: 17.1 m/s, Vertical: 17.1 m/s
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\(A\)

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\(\sin\,40^{\circ}\) \(=\dfrac{v_v}{20}\)  
\(v_v\) \(=20\times \sin\,40^{\circ} = 12.9\ \text{ms}^{-1}\)  
\(\cos\,40^{\circ}\) \(=\dfrac{v_h}{20}\)  
\(v_h\) \(=20 \times \cos\,40^{\circ} = 15.3\ \text{ms}^{-1}\)  

 
\(\Rightarrow A\)

PHYSICS, M1 EQ-Bank 4 MC

A rescue drone is flying with a velocity of 30 m/s, N30°E, relative to the ground. The wind is blowing at 15 m/s in an eastern direction, relative to the ground.

What is the magnitude of the drone’s velocity relative to the wind?

  1. 15.0 m/s
  2. 20.2 m/s
  3. 26 m/s
  4. 33.2 m/s
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\(C\)

Show Worked Solution

Split the vector of the drone into its vector components:

\(v_y= 30\sin 60 = 25.98\ \text{ms}^{-1}\ \text{north}\)

\(v_x = 30\cos 60 = 15\ \text{ms}^{-1}\ \text{east}\)

 
Velocity of the drone relative to the wind:

\(y\text{-component}\ =30\sin\,60^{\circ}=25.98\ \text{ms}^{-1}\ \text{north}\)

\(x\text{-component}\ =30\cos\,60^{\circ}-15 = 0\ \text{ms}^{-1}\ \text{east}\)

\(\therefore\) Velocity of the drone relative to the wind \(= 26\ \text{ms}^{-1}\ \text{north}\)

\(\Rightarrow C\)

PHYSICS, M1 EQ-Bank 8

By separating the vector, shown below, into its constituent horizontal and vertical elements, express each component in terms of theta \((\theta) \).   (2 marks)
 

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Horizontal component: \(v\,\cos \theta\)

Vertical component: \(v\,\sin \theta\)

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Horizontal component: \(v\,\cos \theta\)

Vertical component: \(v\,\sin \theta\)

PHYSICS, M1 EQ-Bank 3

A soccer ball leaves the ground with a velocity of 10 ms\(^{-1}\) at 40° above the horizontal.

  1. Draw a vector diagram to represent the balls initial velocity.   (1 marks)

  2. Calculate, to 2 decimal place, the magnitude of the horizontal and vertical components of the vector.  (2 marks)

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a. 
       

 

b.   
       

\(v_v=6.43\ \text{ms}^{-1}\)

\(v_h=7.66\ \text{ms}^{-1}\)

Show Worked Solution

a. 
       

 

b.   
       

\(v_v=10\sin40^{\circ}=6.43\ \text{ms}^{-1}\)

\(v_h=10\cos40^{\circ}=7.66\ \text{ms}^{-1}\)