In the figure, you throw a ball toward a wall at speed 36.0...

In the figure, you throw a ball toward a wall at speed 36.0 m/s and at angle ?0 = 37.0° above the horizontal. The wall is distance d = 16.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (a) Number 10.55 Unit m (b) Number 28.75 Unit m/s (c) Number 21.67 Unit m/s

Transcript

00:01 So let us start with the concept which we are going to use here for this question.

00:07 That is the second equation of motion says that s is equals to u t plus one half a t square, where u initially speed acceleration t time taken as well with the displacement.

00:18 So according to the question a ball is thrown, let's say from this point, in such a way that it does the projectile motion.

00:28 Now it is thrown at an angle of let's say t 10 not now here, it is nothing but a wall, let's say, which is at a distance of a tree from this point of throwing.

00:42 So we are given the value of d is equals to 16 .0 meters, while angle of throwing is 37 degree and initially speed by which it is thrown is given u is equals to 36 meter per second.

00:59 So we need to calculate in the a part that is how far above the release point does the ball hit the wall.

01:08 So according to newton's second law, we can clearly say that the by applying the second equation of motion, this height h would be equals to the initial velocity that is u times of t plus one half a gt square because acceleration here will be due to gravity.

01:29 Now let us write down the vertical component.

01:32 So let's say v i y is the vertical component of velocity.

01:36 So this will be equals to we initial into sine of theta where i'm considering u is equals to v of i.

01:44 So don't be confused.

01:45 So this is nothing but vertical component substituting this year height h would becomes equals to we have t into sine of theta plus one half gt square.

01:58 Now similarly, the horizontal component of velocity will become v i of x that will be equals to we echo sign of data and that will be equals to the distance covered divided by the time taken...

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