An object is projected from the ground at 32 degrees above the horizontal with a
ID: 1389261 • Letter: A
Question
An object is projected from the ground at 32 degrees above the horizontal with an intial speed of 35 m/s. Assume that there are no frictional forces acting on the object.
a) if the object's mass is 100 kg, determine the constant force that was nessecary to produce the intial velocity if the force acted for 10 ms.
b. determine the time it takes for the object to reach its maximum height.
c.determine the horizontal and verticial components of the velocity when the object is at half of its maximum height and is on the way down.
d. determine the total time that the object is in the air.
e. determine the distance that the object travels while its is in the air.
Please show me the steps and formulas to this problem. I am failing physics and this problem would really help me pass. Please.
Explanation / Answer
a) Initial velocity = 35 m/s
By formula of acceleration, we have a = (v-u)/t,
where a, v, u and t are acceleration, final velocity, initial velocity and time respectively.
Acceleration required for the object to achieve a velocity of 35 from 0 in 10 ms is
a = (35-0)/(10*10-3)
a = 3500 m/s2
Force = mass*acceleration.
So, force required = 100*3500 = 350,000 N
(b) The object reaches maximum height when vertical component of the velocity becomes zero.
Initial vertical component of velocity = 35*sin(32) = 18.55 m/s
Acceleration due to gravity = -9.8 m/s2 (Negative sign is because the force is downwards)
From formula of motion, we have v-u = a*t
where v, u, a and t are final velocity, initial velocity, acceleration and time respectively.
Here, 0 - 18.55 = -9.8*t
t = 1.89 s
Time it takes to reach maximum height = 1.89 seconds
(c) Initial horizontal component of velocity = 35*cos(32) = 29.68 m/s
Horizontal component of velocity remains same over the entire flight time.
So, horizontal component at that instance = 29.68 m/s
Max height is given by H = u2/2g, where u is initial vertical component of velocity(18.55 m/s)
H = 18.55^2 /2*9.8 = 17.55 m
SO, half of max height will be 17.55/2 = 8.78 m/s
By equation of motion, v2 - u2 = 2as
where v, u, a and s are final velocity, initial velocity, acceleration and displacement respectively.
Here v^2 - 18.55^2 = 2*(-9.8)*8.78
v^2 = 172.1
v = 13.11 m/s or -13.11 m/s
Since object is travelling down, vertical component of velocity = -13.11 m/s
(d) TIme of flight = 2u*sin(theta)/g
where u, theta are initial velocity and angle of projection respectively.
So, t = 2*35*sin(32)/9.8
= 3.79 seconds
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