What is the range of the function y=4e^x
2 answers:
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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
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
⇒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 .
=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
⇒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.