Explanation:
Given that,
Horizontal velocity = 25 m/s
Time = 3.00 sec
(I). We need to calculate the horizontal range
Using formula of range
[tex]R=v\times t[/tex]
Put the value into the formula
[tex]R=25\times3.00[/tex]
[tex]R=75\ m[/tex]
The horizontal range is 75 m.
(II). We need to calculate the initial vertical component of velocity
Using equation of motion
[tex]v=u+gt[/tex]
Where, v = final velocity
u = initial velocity
g = acceleration due to gravity
t = time
Put the value into the formula
[tex]0=u-9.8\times1.5[/tex]
[tex]u=14.7\ m/s[/tex]
The initial vertical component of velocity is 14.7 m/s.
(III). We need to calculate the initial angle of projection
Using formula of angle
[tex]\tan\theta=\dfrac{vertical\ velocity}{horizontal\ velocity}[/tex]
[tex]\theta=\tan^{-1}\dfrac{14.7}{25}[/tex]
[tex]\theta=30.45^{\circ}[/tex]
The initial angle of projection is 30.45°.
Hence, This is the required solution.
