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PHYSICS, M5 2015 HSC 20 MC

A projectile was launched from the ground. It had a range of 70 metres and was in the air for 3.5 seconds.

At what angle to the horizontal was it launched?

  1. 30°
  2. 40°
  3. 50°
  4. 60°
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`B`

Show Worked Solution

Find `u_(x):`

`s_(x)` `=u_(x)t`  
`u_(x)` `=(70)/(3.5)`  
  `=20  text{ms}^(-1)`  

 
Find `u_(y)` (projectile has a vertical velocity of zero at its maximum height):

`v_(y)` `=u_(y)+a_(y)t`  
`0` `=u_(y)-9.8xx1.75`  
`u_(y)` `=17.15  text{ms}^(-1)`  

  
Find launch angle:

`tan theta` `=(u_(y))/(u_(x))`  
`theta` `=tan^(-1)((17.15)/(20))`  
  `=40^(@)`  

 
`=>B`


♦ Mean mark 55%.

PHYSICS, M5 2020 HSC 24

The graph shows the vertical displacement of a projectile throughout its trajectory. The range of the projectile is 130 m.
 

Calculate the initial velocity of the projectile.   (4 marks)

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37  m `text{s}^(-1)`, at 54° above the horizontal.

Show Worked Solution

From the graph, at `t=3`,  the projectile reaches a maximum height of 44 m:

`s_(y)` `=u_(y)t+(1)/(2)a_(y)t^(2)`  
`44` `=u_(y)(3)-(1)/(2)(9.8)(3^(2))`  
`u_(y)` `=29.4` m `text {s}^(-1)`  

  
Find `u_x` given time of flight = 6 s:

`u_(x)` `=(s_(x))/(t)`  
  `=(130)/(6)`  
  `=21.7` m `text{s}^(-1)`  

   
Using Pythagoras:

`u^(2)` `=u_(x)^(2)+u_(y)^(2)`  
  `=21.7^(2)+29.4^(2)`  
`u` `=37` m `text{s}^(-1)`  

  
Find launch angle (`theta)`:

`tan theta` `=(u_y)/(u_x)`  
`theta` `=54^(@)`  

  
So,  `u=37` m `text{s}^(-1)`, at 54° above the horizontal.