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A projectile is shot from the ground at an angle of 55 degrees with respect to the horizontal, and it lands on the ground 5 seconds later. Find: (a) (10 pts) the horizontal component and the vertical component of initial velocity; (b) (10 pts) the magnitude and the direction of velocity when the projectile hits the ground.
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Added Wed, 08 Jul '15
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1. The horizontal and the vertical motion can be treated separately.

2. The speed in the horizontal dirction remains constant an dequal to the speed during release.

3. The time of flight is dependent on the vertical component of the initial velocity.

Now the whole process of projectile can be broken down as 2 processes. One where you are throwing a ball directly overhead in the air and secondly u are rolling the ball on the ground.The throwing part will give it the height and the time of flight whereas the rolling part will give it the range.

Since the projectile is falling 5 seconds later, it means the time of flight is 5 seconds. T=2uvert/g where u = initial velocity and g = acc. due to gravity. and uvert is the vertical component of the initial velocity u. This gives uvert= 2.5g . Assuming g=9.81 m/s^2, we have uvert=24.525 m/s^2. \theta= angle of projection = 55 degrees . Hence the horizontal component of initial velocity u given as uhor can be given by the equation tan\theta =uvert / uhor Which gives uhor as 24.524/tan 55 =17.17 m/s^2

Hence uhor=17.17 m/s^2, uvert=24.525 m/s^2 which answers the first part of the question.

For the second part, we can deduce that the horizontal component of velocity will remain the same as no external force is acting on the ball in horizontal direction. So uhor=vhor where v is the final velocity. We can also argue that the gravitational force that decelaretes he body in the ascent accelerates it during descent and hence the net vertical speed remains the same. So the magnitude of initial velocity and final velocity remains the same. v=( vhor^2 +vvert ^2)1/2 = 29.93 m/s^2.

However the direction will change as the direction of vertical velocity has reversed. The new direction will be 55 degrees into the ground or 360-55=305 degrees in anticlockwise direction wrt to the ground as the projectile is symmetrical.


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In other words : 

time of flight =2vsin(55)/g =5

v=29.914 m/sec

so horizontal component =29.914 cos(55) =17.157 m/sec (ans)

vertical component =29.914 sin(55) =24.504 m/sec (ans)

b)magnitude and the direction of velocity when the projectile hits the ground.=29.914 at an angle of 55 degrees with respect to horizontal downwards
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