Question

The position coordinates of a projectile thrown from
ground are given by y = 3t - 5t(m) and x = 4t (m)
(here t is in second, x is horizontal and y is vertical)
horizontal range of the projectile is
(1) 2.4 m
(2) 4.8 m
(3) 9.6 m
(4) 42 m​

Answer 1

Answer:

The position coordinates of a projectile thrown from

ground are given by y = 3t - 5t(m) and x = 4t (m)

(here t is in second, x is horizontal and y is vertical)

horizontal range of the projectile is

(1) 2.4 m

(2) 4.8 m

(3) 9.6 m

(4) 42 m

Answer 2

The horizontal range of the projectile is equal to 2.4 metres. Hence, the correct option is (1) 2.4 m

Given:

The function of x-coordinate and time (t): x = 4t

The function of y-coordinate and time (t): y = 3t - 5t²

To Find:

The horizontal range of the projectile (R).

Solution:

→ When a particle is projected from the horizontal with a velocity 'v' making an angle 'θ' then the particle's velocity has two components: The vertical component and the horizontal component. The vertical component of the particle's velocity is equal to vsinθ and the horizontal component of the particle's velocity is equal to vcosθ.

→ To calculate the range of the projectile (R), we will have to calculate the x-coordinate of the endpoint of the projectile.

→ The vertical height that is the y-coordinate is zero at the starting point and the endpoint of the projectile.

∴ Putting the value of 'y' as zero.

∴ y = 3t - 5t² = 0

∴ t (3 - 5t) = 0

∴ t = 0 or 3/5 seconds.

→ Therefore, it can be said that the particle hits the endpoint of the projectile at t = 3/5 seconds.

→ Now calculating the value of the x-coordinate at t = 3/5 seconds:

∴ x = 4t

∴ x = 4(3/5)

∴ x = 12/5

∴ x = 2.4 metres

∵ The x-coordinate at t = 3/5 seconds is equal to 2.4 metres.

Therefore the horizontal range (R) of the projectile is equal to 2.4 metres. Hence, the correct option is (1) 2.4 m

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