A ball is thrown horizontally from the top of a 525 m...

A ball is thrown horizontally from the top of a 52.5-m building and lands 122.3 m from the base of the building. Ignore air resistance. (Assume the ball is thrown in the +x direction and upward to be in the +y direction.) (a) How long (in s) is the ball in the air? s (b) What must have been the initial horizontal component of the velocity (in m/s)? (Indicate the direction with the sign of your answer.) m/s (c) What is the vertical component of the velocity (in m/s) just before the ball hits the ground? (Indicate the direction with the sign of your answer.) m/s (d) What is the velocity (in m/s) (including both the horizontal and vertical components) of the ball just before it hits the ground? ____ magnitude m/s direction ° ____counterclockwise from the +x-axis

Transcript

00:01 The horizontal distance needed to be covered by the projectile is 52 .5 meters.

00:11 While the horizontal distance is sx, the vertical distance is s -y, the horizontal distance covered by the projectile will be 122 .3 meters.

00:27 Now, solving for the first part, we need to determine how long the ball will be in here.

00:35 Let us apply the second kinematical equation.

00:41 Sy is equal to uyt plus half ay t square in vertical direction.

00:49 We know that initial velocity in vertical direction was zero because the projectile was launched horizontally and there was no vertical component.

00:58 Therefore t can be determined by 2, sy divided by ay.

01:06 The acceleration will be, the acceleration for the object coming downward will be, of course, now the distance traveled is 52 .5 meters.

01:25 The acceleration, ay, is g whose value is 9 .8 meter per second square, and the time will come out to be equals to 3 .3 .27.

01:43 This is the answer for part a.

01:49 For solving part b, we need to determine the x component of initial velocity.

01:57 Now we know that sx is equal to uxt since the acceleration is 0.

02:05 This is a result from second kinematical equation in horizontal direction.

02:11 Now ux can be determined as sx.

02:16 By t we know the value of s x is 122 .3 meters and the value of t is 3 .27 seconds we get the value of ux2b equals to 37 .40 meter per second this is the answer for part b of course the direction must be positive since the ball is thrown in positive x direction the initial velocity was all positive part c, it is asking the vertical component of y, a vertical component of the final velocity, when the ball hits the ground, that is the final velocity, we will use the first kinematical equation here which says vy is equals to uy plus gt.

03:14 This is the equation in vertical direction.

03:19 Now, substituting the value of uy, we know that there was no initial velocity along y direction.

03:30 All the velocity was in x direction.

03:33 There was no velocity initially in y direction...

Question

Please give Ace some feedback