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A projectile fired up into the air at an angle has a range of 235 m and a flight time of 47 s.
a. what is the horizontal component of the projectiles velocity?
b. what is the maximum height of the projectile?
c. calculate viy for the projectile

Respuesta :

Answer:5[tex]ms^{-1}[/tex],133.6[tex]m[/tex],51.18[tex]ms^{-1}[/tex]

Explanation:

Let [tex]v_{x}[/tex],[tex]v_{y}[/tex] be the horizontal and vertical components of velocity.

Question a:

Horizontal component of velocity is the ratio of range and time of flight.

So,horizontal component of velocity is [tex]\frac{235}{47}=5ms^{-1}[/tex]

So,[tex]v_{x}=5ms^{-1}[/tex]

Question b:

Time of flight=[tex]\frac{2v_{y}}{g}[/tex]

So,[tex]v_{y}=\frac{47\times 9.8}{2}=51.18ms^{-1}[/tex]

Maximum height is given by [tex]\frac{v_{y}^{2}}{2g}[/tex]

So,maximum height is [tex]\frac{51.18^{2}}{2\times 9.8}=133.6m[/tex]

Question c:

The vertical velocity is already calculated in Question b.

[tex]v_{y}=51.18ms^{-1}[/tex]

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