Suppose the ball is thrown from the same height as in the...

Suppose the ball is thrown from the same height as in the PRACTICE IT (h = 48.0 m) problem at an angle of 29.0° below the horizontal. If it strikes the ground 56.3 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.) (a) the time of flight (s) (b) the initial speed (m/s) (c) the speed and angle of the velocity vector with respect to the horizontal at impact speed (m/s) angle (° below the horizontal)

Transcript

00:01 We'd like to use our equation to find the time of flight, the initial speed, and the speed and angle.

00:07 Okay, so let's use our first, our kinematic equation of motion.

00:12 We have our initial velocity in the y direction is v .0 sine of 29 degrees times time plus 4 .9 times squared.

00:23 Now our horizontal displacement is v .0 cosine of 29 degrees times times.

00:31 T so um and that is equal to our 56 .3 meters as given in our problem now we can use this so we have 48 meters is equal to our 56 .3 divided by t cosine of 29 degrees times sine of 29 degrees times t plus 4 .9 t squared we then get 48 meters is equal to 56 .3 times tangent of 29 degrees plus 4 .9 t squared.

01:13 And then this is just going to give us, because our t's canceled out, we can then just use this to solve for t.

01:21 So we have 48 is equal to 56 .3 tangent of 29 degrees times or plus 4 .9 t squared.

01:32 We then use this to solve for our t.

01:39 So taking our 56 .3 tangent of 29 degrees, that's 31 .2.

01:47 And so we get t to be equal to 1 .85 seconds.

01:52 So that's going to be our answer for our time of flight.

01:56 Now for b, we'd like to find the initial speed.

01:59 The initial is 56 .3 divided by t cosine of 29 degrees.

02:07 So it'll be our 56 .3 divided by 1 .85 cosine of 29 degrees to get 34 .7 meters per second as our initial speed...