An object is projected from the ground at 32 degrees above the horizontal with a
ID: 1521178 • Letter: A
Question
An object is projected from the ground at 32 degrees above the horizontal with an initial speed of 35 m/s. Assume that there are no frictional forces acting on the object. If the objectprimes mass is 100 kg. determine the constant force that was necessary to produce the initial velocity if the force acted for 10 ms. Determine the time it takes for the object to reach its maximum height. Determine the horizontal and vertical components of the velocity when the object is at half of its maximum height and is on the way down. Determine the total time that the object is in the air. Determine the distance that the object travels while it is in the air.Explanation / Answer
angle =32
initial velocity u=35 m/s
(a) ans
we find the force
the force F= mucos(theta)/t
=100*35 *cos(32)/10*10^-3
=2.97*10^5 N
therefore the force applied F=2.97*10^5 N
(b) ans
the body to reach its maximum height
maximum height H max =U^2 sin^2 (theta)/2g
=(35)^2 sin^2(32)/2*9.8
=17.6 m
now we find time to reach maximum height
using the kineticmatices equation
s=ut+1/2gt^2
Hmax=usin(theta)t-1/2gt^2
17.6 =35 sin(32) t - 0.5*9.8 *t^2
=>4.9 t^2-18.5 t+17.6=0...........................................(1)
therefore the quartic equation ax^2+bx+c=0...................................(2)
use the equation x=-b+_[b^2 -4ac]^½/2a
compare eq 1 &2 we find time using above equation
time t=-(-18.5)+[18.5^2-4*4.9*17.6]^½/[2*4.9]
=18.5-1.65/9.8
=1.72 sec
(c) ans
the horizantal velocity Ux=u cos(theta)
=35*cos(32)
=29.7 m/s
the vertical velocity Uy =u sin(theta)
=35 sin(32)
=18.55 m/s
(d) ans
therefore the time of flight T=2u sin(theta)/g
the time of flight => T=2*35*sin(32)/9.8
T=3.79 sec
(e) ans
the distance of object travel while is in air
vertical distance =u^2 sin^2(theta)/2g
=35^2 sin^2(32)/2*9.8
=17.6 m
the horizantal distane =u^2 sin 2(theta)/g
=35^2 sin 2(32)/9.8
=112.35 m