Respuesta :
Answer:
(a) The velocity after 2 second is 5.9 m/s
The velocity after 4 second is -13.7 m/s.
(b) The projectile reaches its maximum height after 2.60 s of projection.
(c)The maximum height that is attained by the projectile is 37.18 m.
(d)Therefore the projectile hits the ground after 5.36 seconds of projection.
(e)The velocity of the projectile when it hits the ground is 27.03 m/s
Step-by-step explanation:
Given that, a projectile shot vertically upward from a point 4 m above the ground with a initial velocity of 25.5 m/s.
The height of the projectile after t seconds is
[tex]h=4+25.5t-4.9t^2[/tex]
where h is in meter.
(a)
We use the formula
v=u+at
V= final velocity
u = initial velocity = 25.5 m/s
a = acceleration= acceleration due to gravity= 9.8 m/s²
Since the object moves upward direction and acceleration due to gravity is downward direction. So here a= -9.8 m/s.
v(2)= 25.5+(-9.8)×2
=25.5-19.6
=5.9 m/s
And when t= 4
v(4)= 25.5+(-9.8)×4
=25.5-39.2
= -13.7 m/s
The velocity after 2 second is 5.9 m/s
The velocity after 4 second is -13.7 m/s.
(b)
At its maximum height,the velocity of the projectile is zero. i.e v=0
∴0=25.5+(-9.8)t
⇒9.8t=25.5
[tex]\Rightarrow t=\frac{25.5}{9.8}[/tex]
⇒t = 2.60 s
The projectile reaches its maximum height 2.60 s after projection.
(c)
To find the maximum height, we are putting t= 2.60 in this equation [tex]h=4+25.5t-4.9t^2[/tex].
[tex]\therefore h= 4+(25.5\times 2.60)-(4.9\times 2.60^2)[/tex]
=37.18 m
The maximum height that is attained by the projectile is 37.18 m.
(d)
When the projectile hits the ground the height will be zero i.e h=0
From the equation of height we get
[tex]\therefore h=0=4+25.5t-4.9t^2[/tex]
[tex]\Rightarrow 4+25.5t-4.9t^2=0[/tex]
[tex]\Rightarrow t=\frac{-25.5\pm\sqrt{25.5^2-4(-4.9).4}}{2(-4.9)}[/tex]
⇒t= -0.15 ,5.36
Therefore it hits the ground after 5.36 seconds of projection.
(e)
To find the velocity we use the formula v=u+at
Here v = final velocity=?
u=25.5 m/s,
t = 5.36 s
a= -9.8m/s²
v=25.5+(-9.8)5.36
= -27.03 m/s
Negative sign denoted that the motion of the projectile is downward direction.
The velocity of the projectile when it hits the ground is 27.03 m/s.
(a) The velocity of the projectile after 2 s and 4 s is 5.9 m/s and -13.7 m/s respectively.
(b) the projectile will reach maximum height in 2.6 s.
(c) the maximum height of the projectile is 37.18 m
(d) The time for the projectile to fall to the ground is 2.76 s
(e) The velocity with which the projectile hits the ground is 27.1 m/s.
The given equation;
[tex]h = 4 + 25.5t - 4.9t^2[/tex]
(a) the velocity after 2 s and 4 s
[tex]v = \frac{dh}{dt} = 25.5 - 9.8t\\\\v(2) = 25.5 - 9.8(2)\\\\v(2) = 5.9 \ m/s\\\\v(4) = 25.5 - 9.8(4)\\\\v(4) = -13.7 \ m/s[/tex]
(b) when the projectile reaches maximum height the final velocity is zero.
[tex]v = \frac{dh}{dt} = 0\\\\\frac{dh}{dt} = 25.5 - 9.8t = 0\\\\9.8t = 25.5\\\\t= \frac{25.5}{9.8} = 2.6 \ s[/tex]
Thus, the projectile will reach maximum height in 2.6 s.
(c) the maximum height of the projectile is calculated as;
[tex]h = 4 + 25.5t - 4.9t^2\\\\h(2.6) = 4 + 25.5(2.6) - 4.9(2.6^2)\\\\h(2.6) = 37.18 \ m[/tex]
(d) The time for the projectile to fall to the ground is calculated as;
[tex]h = v_y_0 t + \frac{1}{2} gt^2\\\\37.18 = 0 + 0.5\times 9.8t^2\\\\37.18 = 4.9t^2\\\\t^2 = \frac{37.18}{4.9} \\\\t^2 = 7.588\\\\t = \sqrt{7.588} \\\\t = 2.76 \ s[/tex]
(e) when the projectile hits the ground, the velocity will be maximum;
[tex]v'' =\frac{d^2h}{dt^2} = -9.8[/tex]
[tex]v_f= v_0 + gt\\\\v_f = 0 + 9.8\times 2.76\\\\v_f = 27.1 \ m/s[/tex]
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