Answer :
(a)The initial speed of the projectile 28.19 m/s
(b)The total time interval the projectile was in flight 4.068 s
(c)Range reduces and flying time rises as the angle increases.
What is a time interval?
A larger span of time can be broken up into several shorter, equal-length segments. These are referred to as time periods. Consider the scenario if you wished to gauge a car's speed throughout an hour-long trip. You may break up an hour into ten-minute halves.
Briefing:
(a)The equation: gives the projectile's range.
[tex]$d=\frac{u^2 \sin 2 \theta}{g}$[/tex]
where
u is the initial speed
0 is the angle of projection
g = 9.8 m/s2 is the acceleration of gravity
In relation to the projectile in this issue, we are aware of:
d = 81.1 m (range)
θ = 45⁰
Therefore, by changing the equation, we can get the beginning speed:
[tex]$u=\sqrt{\frac{d g}{\sin 2 \theta}}=\sqrt{\frac{(81.1)(9.8)}{\sin (2-45)}}=28.19 \mathrm{~m} / \mathrm{s}$[/tex]
(b)
The projectile's initial vertical velocity is
[tex]$u_y=u \sin \theta$[/tex]
and since we are aware that the vertical velocity zeroes out at the trajectory's highest point, we have
[tex]$v_y=u_y-g t=0$[/tex]
This indicates that the amount of time required to reach the highest point is
[tex]$t=\frac{u_y}{g}$[/tex]
Actually, the flight takes twice as long as this, thus
[tex]$t=\frac{2 u \sin \theta}{g}$[/tex]
And by substituting
u = 28.19 m/s
θ = 45⁰
g = 9.8 m/s²
we find:
[tex]$t=\frac{2(28.19) \sin 45^{\circ}}{9.8}=4.068 \mathrm{~s}$[/tex]
(c)
Range reduces and flying time rises as the angle increases.
In order to respond, let's examine the two formulas:
[tex]$d=\frac{u^2 \sin 2 \theta}{g}$[/tex]
[tex]$t=\frac{2 u s i n \theta}{g}$[/tex]
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