There are 438 students at a middle school. If 38% of the students participate in after school sports what would the number of st
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(a) The projectile remains in the air for 10 seconds
(b) It travels a horizontal distance of 600 meters
Step-by-step explanation:
A projectile is fired at such an angle that
1. The vertical component of its velocity is 49 m/sec
2. The horizontal component of its velocity is 60 m/sec
We need to find:
(a) How long the projectile remains in the air
(b) The horizontal distance it travels
∵ The vertical distance y = t -
g t², where
is the vertical component of its velocity, g is the acceleration
of gravity and t is the time
∵ y = 0 ⇒ it return to the same initial height
∵ = 49 m/s
∵ g = 9.8 m/s²
- Substitute these values in the rule above
∴ 0 = 49 t - (9.8) t²
∴ 0 = 49 t - 4.9 t²
- Take t as a common factor
∴ 0 = t (49 - 4.9 t)
- Equate each term by 0
∴ t = 0 ⇒ at initial position
∴ 49 - 4.9 t = 0
- Add 4.9 t to both side
∴ 49 = 4.9 t
- Divide both sides by 4.9
∴ t = 10 seconds
(a) The projectile remains in the air for 10 seconds
∵ The horizontal distance = t, where
is the
horizontal component of its velocity
∵ = 60 m/sec
∵ t = 10 seconds
∴ x = 60 × 10 = 600 meters
(b) It travels a horizontal distance of 600 meters
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Answer:
Anish performed better than Aman.
Step-by-step explanation:
Given:
Marks obtained by Aman =410
Total marks = 500
We will first find the percentage of marks obtained.
To find the percentage of marks obtained we will divide Marks obtained by Total marks multiplied by 100.
framing in equation form we get;
Percentage of Aman =
Also Given:
Marks obtained by Anish = 536
Total marks = 600
We will first find the percentage of marks obtained.
To find the percentage of marks obtained we will divide Marks obtained by Total marks multiplied by 100.
framing in equation form we get;
Percentage of Anish =
Now we can see;
Percentage of Anish is greater than Percentage of Aman.
Hence Anish performed better than Aman.