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E XAMPLE P ROBLEMS Projectile Motion. A BALL IS THROWN HORIZONTALLY FROM THE TOP OF A BUILDING 47.7 M HIGH. T HE BALL STRIKES THE GROUND AT A POINT 108.

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Presentation on theme: "E XAMPLE P ROBLEMS Projectile Motion. A BALL IS THROWN HORIZONTALLY FROM THE TOP OF A BUILDING 47.7 M HIGH. T HE BALL STRIKES THE GROUND AT A POINT 108."— Presentation transcript:

1 E XAMPLE P ROBLEMS Projectile Motion

2 A BALL IS THROWN HORIZONTALLY FROM THE TOP OF A BUILDING 47.7 M HIGH. T HE BALL STRIKES THE GROUND AT A POINT 108 M FROM THE BASE OF THE BUILDING. A. Find the time the ball is in motion. B. Find the initial velocity C. Find the x component of the velocity just before it hits the ground D. Find the y component of the velocity just before it hits the ground

3 T HE TIME THE BALL IS IN THE AIR Is gravity effecting it? Yes, use the vertical equations. D=1/2gt 2 √2d/g 2*47.7/9.8 √9.7=3.12s

4 I NITIAL V ELOCITY OF THE BALL V=d/t 108/3.12 V=34.6

5 F IND THE X COMPONENT OF THE VELOCITY JUST BEFORE IT HITS THE GROUND The same as the initial velocity

6 F IND THE Y COMPONENT OF THE VELOCITY JUST BEFORE IT HITS THE GROUND V=gt V=-9.8*3.12 V= m/s ***Remember it is falling down and gravity would be negative

7 A N ARROW IS SHOT AT A 33 ⁰ ANGLE WITH THE HORIZONTAL. I T HAS A VELOCITY OF 54 M / S A. How high will it go? B. What horizontal distance will it travel?

8 54m/s 33 ⁰ v y V x Sin 33 =y/54 Sin 33 *54 =29.41 m/s Cos 33=x/54 Cos 33 * 54= 45.28m/s Vertical speed Horizontal speed

9 Look at your vertical equations D=1/2gt 2 V=gt Let’s combine these two equations using substitution method. t=v/g d=1/2 g(v/g) 2 d=44.13 m

10 N OW LETS SOLVE FOR THE HORIZONTAL DISTANCE. We solved for the horizontal velocity and now we need to find an equation for the horizontal distance. D=vt and because of the trajectory we need to use twice the time from the previous problem.

11 A DESCENT VEHICLE LANDING ON THE MOON HAS A VERTICAL VELOCITY VECTOR TOWARD THE SURFACE OF THE MOON OF 31.6 M / S. A T THE SAME TIME, IT HAS A HORIZONTAL VELOCITY OF 53.2 M / S. At what speed does the vehicle move along its descent path? At what angle with the vertical is its path?

12 A T WHAT SPEED DOES THE VEHICLE MOVE ALONG ITS DESCENT PATH ? 31.6 m/s R 53.6 m/s a 2 +b 2 =c 2

13 A T WHAT ANGLE WITH THE VERTICAL IS ITS PATH ? 31.6 m/s R θ=? 53.6 m/s tan θ=o/a Θ=tan -1 (o/a)

14 J ANET JUMPS OFF A HIGH DIVING PLATFORM WITH A HORIZONTAL VELOCITY OF 2.83 M / S AND LANDS IN THE WATER 2.2 S LATER. A. How high is the platform? B. How far from the base of the platform does she land?

15 A IS ASKING FOR THE VERTICAL DISTANCE D=1/2gt 2

16 B IS ASKING FOR HORIZONTAL DISTANCE D=vt


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