Answer:
A:7.2[tex]ms^{-1}[/tex]
B:14.25[tex]ms^{-1}[/tex]
C:1.45[tex]sec[/tex]
D:10.3[tex]m[/tex]
E:2.9[tex]sec[/tex]
F:20.88[tex]m[/tex]
Explanation:
Let [tex]v[/tex] be the velocity and [tex]\alpha[/tex] be the angle between the velocity and ground.
Question A:
Horizontal component of velocity is given by [tex]vcos(\alpha )[/tex].
So,horizontal component is [tex]16\times cos(63)=16\times 0.45=7.2ms^{-1}[/tex]
Question B:
Vertical component of velocity is given by [tex]vsin(\alpha )[/tex].
So,vertical component is [tex]16\times sin(63)=16\times 0.89=14.25ms^{-1}[/tex]
Question C:
Time required is given by [tex]\frac{\text{vertical component of velocity}}{g}}=\frac{14.25}{9.8}=1.45 seconds[/tex]
Question D:
Maximum height is given by [tex]\frac{\text{vertical component of velocity}^{2}}{2g}}=\frac{203.06}{19.6}=10.3m[/tex]
Question E:
Time of flight is twice the time required to reach maximum height=[tex]2\times 1.45=2.9 seconds[/tex].
Question F:
The distance between the player and ball after landing is called range and is given by [tex]\text{horizontal component of velocity}\times \text{time of flight}=[/tex][tex]7.2\times 2.9=20.88m[/tex]
