Maximum Height Of A Projectile

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Strings (SiPjAjk) = S7P4A42     Base Sequence = 12735     String Sequence = 12735 - 4 - 42



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Maximum Height Of A Projectile
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The parametrc equatons of the motion of a projectile are as follows:
x = 20t ------(1)
y = -16t2 + 30t -------(2)
(a) What is the maximum height of the projectile?
(b) What is the initial velocity of the projectile?
(c) What is the velocity of the projectile when it hits the ground?
(d) What is the range of the projectile?

The string:
(a) S7P4A42 (Motion - Parabolic).
(b) and (c) S7P5A51 (Physical Change - Velocity)
(d) S7P4A41 (Linear Motion)
The math:
Pj Problem of Interest (PPI) is of type motion. Problems of distance traveled are motion problems.
Vertical velocity of projectile, v = 30 - 32t ---------(3) (derivative of eqn(2))
The projectile attains its maximum height when its vertical velocity, v = 0.
v = 0, when t = 30/32 sec (from eqn(3))
substitute t = 30/32 in eqn (2) to get maximum height of projectile
So, maximum height = 30(30/32) - 16(30/32)2 = 14.1 ft
(b) Pj Problem Of Interest (PPI) is of type change. Speed, velocity, acceleration and t (duration) problems are change problems.
Initial velocity of projectile = 30 ft/sec
(c) Projectile hits the ground when y = 0.
When y = 0, t = 15/8 (from eqn (2)).
Subtituting t = 15/8 in equation (3), we have:
Velociy of projectile when it hits the ground = 30 - 32(15/8) = 30 - 60 = -30 ft/sec.
(d) Pj Problem Of Interest (PPI) is of type motion. Distances due to motion are motion problems.
The range of a projectile is the horizontal distance from the origin of the projectile to the point when the projectile hits the ground.
So, substituting t = 15/8 in equation (1), we have:
20t =20(15/8) = 75/2 = 37.5
So, range of projectile is 37.5 ft.

The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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