Question

a projectile is lunched with an initial speed of 60.0mm/s at an angle of 30.0° above the horizontal.The projectile lands on a hillside 4 s later.Neglect air friction.(a)what is the projectile's velocity at the highest point of its trajectory?​

Answer 1

Answer:

52 mm/s (approximately)

Explanation:

Given:

Initial speed of the projectile is, $u=60.0\ mm/s$

Angle of projection is, $\theta=30.0\°$

Time taken to land on the hill is, $t=4\ s$

In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.

So, the velocity in the horizontal direction always remains the same.

The horizontal component of initial velocity is given as:

$u_x=u\cos\theta\\u_x=60\times \cos(30)\\u_x=30\sqrt3\approx52\ mm/s$

Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.

Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.