The range of the projectile is maximum at angle 45°, the range is 10.2m, the time to reach the maximum height is 0.72 s and the maximum height is 2.6 m
The stone, or object or anything projected in a trajectory path is known as a projectile.
In the given two-dimensional projectile motion problem involving a ball starting and ending at the same height. the projectile leaves the ground with an initial velocity of 10 m/s at some angle Ф with the horizontal,
(a) The value of the angle Ф (in degrees) which will makes the range of the projectile maximum is angle 45° because Range = u²sin2Ф ÷ g
At Ф = 45
Range = u²sin90° ÷ g
where sin 90 = 1
So, Range = u² / g which is the maximum range.
(b) The range at this angle will be
Range = 10² ÷ 9.8
Range = 100 / 9.8
Range = 10.2 m
(c) To calculate how long it takes the ball to reach maximum height, we will use the formula
t = usinФ ÷ g
t = 10 sin 45 ÷ 9.8
t = 0.72s
(d) The maximum height will be
H = u²sin²Ф ÷ 2g
H = 10²(sin45)² ÷ 2 × 9.8
H = 100 × 0.5 ÷ 19.6
H = 50 ÷ 19.6
H = 2.55 m
Therefore, the range of the projectile is maximum at angle 45°, the range is 10.2m, the time to reach the maximum height is 0.72 s and the maximum height is 2.6 m
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