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Motion in a Plane

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Question : 32
Total: 32
a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by θ(t)=tan−1(
voy−gt
vox
)
(b) Shows that the projection angle θ0 for a projectile launched from the origin isgiven by θ0=tan−1(
4hm
R
)
where the symbols have their usual meaning.
Solution:  
(a) Let vox and voy be the initial component velocity of the projectile at O along OX direction and OY direction respectively, where OX is horizontal and the OY is vertical. Let the projectilego from O to P in time t and vx,vy be the component velocity of projectile at P along horizontal and vertical directions respectively. Then, vy=voy–gt and vx=vox
If θ is the angle which the resultant velocity
→
v
 makeswith horizontal direction, then
tan‌θ=
vy
vx
=
voy−gt
vox
 or θ=tan−1(
voy−gt
vox
)
(b) In angular projection,
Maximum vertical height, hm=
u2sin2θ0
2g

Horizontal range,R=
u2‌sin‌2‌θ0
g
=
u2
g
2‌sin‌θ0‌cos‌θ0
So,
hm
R
=
tan‌θ0
4
 or tan‌θ0=
4hm
R
 or θ0=tan−1(
4hm
R
)
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