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  • A projectile is launched from the foot of an inclined plane which makes an angle of 30 degrees with the horizontal. The projectile's initial velocity is 20 m/s at an angle of 45 degre

A projectile is launched from the foot of an inclined plane which makes an angle of 30 degrees with the horizontal. The projectile's initial velocity is 20 m/s at an angle of 45 degrees with the inclined plane. Neglecting air resistance, the time taken by the projectile to hit the inclined plane is closest to:

Option: 1

 1.0 s


Option: 2

1.5 s


Option: 3

2.0 s

 


Option: 4

2.5 s


Answers (1)

best_answer

The projectile's motion can be divided into two parts:

Horizontal motion and vertical motion.

In the horizontal direction, the projectile moves with a constant velocity of 20cos(45) = 14.14 m/s.

In the vertical direction, the projectile experiences a constant acceleration due to gravity of 9.8 m/s^2.

Let's consider the vertical motion of the projectile.

The initial vertical velocity of the projectile is 20sin(45) = 14.14 m/s.

The time taken by the projectile to hit the inclined plane can be found using the equation:   

y=v_it + \frac{1}{2}a\times t^{2}

where y is the vertical displacement of the projectile, v_i is the initial vertical velocity of the projectile, a is the acceleration due to gravity, and t is the time taken by the projectile to hit the inclined plane.

The vertical displacement of the projectile can be found using the angle of the inclined plane:

 y=x\tan 30

where x is the horizontal displacement of the projectile.

The horizontal displacement of the projectile can be found using the time taken by the projectile to hit the inclined plane and the horizontal velocity of the projectile:

x = 14.14\times cos(45)\times t

Substituting these equations into the first equation, we get

14.14\tan (30) t = 14.14t\cos 45t+\frac{1}{29.8}\times t^{2}  

Simplifying and solving for t, we get:

t \approx 1.5 s

 

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