Example Problems for Motion in 2-d Answers 1. A stone is thrown horizontally at 15 m/s from a cliff 44 m high. (a) How long does it take the stone to reach the ground? vx = 15 m/s Stone takes parabolic path to the ground Gravity acts in a vertical direction; horizontal velocity as no effect on gravity 44 m B A So gravity acts equally on object projected horizontally (which lands at A), and object that is dropped (which lands at B) To find time of flight to A (which is difficult), find instead the time it takes to drop the same vertical distance to B (easier)

t = 3.0 s (a) t = ? Vertically, vi = 0 g = - 9.8 m/s2 d = - 44 m vx = 15 m/s g = - 9.8 m/s2 d = - 44 m Use d = vi t + ½ g t2 44 m d = 0 + ½ g t2 B A 2 ( d = ½ g t2 ) 2 2 ( - 44 m ) 2 d = g t2 2 d t = = - 9.8 m/s2 g g g 2 d t2 = g t = 3.0 s

(b) How far from the base of the cliff does it land? vx = 15 m/s t = 3.0 s Gravity acts vertically; horizontal motion is unaffected 44 m dx = ? Motion in horizontal is at constant velocity dx = vx t = ( 15 m/s )( 3.0 s ) dx = 45 m

2. A stone is thrown horizontally at 10.0 m/s from a cliff 78.4 m high. (a) How long does it take the stone to reach the ground? vx = 10.0 m/s Instead of finding time of flight to A, find the time it takes to drop the same vertical distance to B 78.4 m B A

t = 4.0 s (a) t = ? Vertically, vi = 0 g = - 9.8 m/s2 d = - 78.4 m vx = 10.0 m/s g = - 9.8 m/s2 d = - 78.4 m Use d = vi t + ½ g t2 78.4 m d = 0 + ½ g t2 B A 2 ( d = ½ g t2 ) 2 2 ( - 78.4 m ) 2 d = g t2 2 d t = = - 9.8 m/s2 g g g 2 d t2 = t = 4.0 s g

(b) How far from the base of the cliff does it land? vx = 10.0 m/s t = 4.0 s Gravity acts vertically; horizontal motion is unaffected 78.4 m dx = ? Motion in horizontal is at constant velocity dx = vx t = ( 10.0 m/s )( 4.0 s ) dx = 40.0 m

3. vx = 0.800 m/s Height of table = 0.960 m dx = vx t t = ? Use dy = viy t + ½ g t2 dx = ? 2 dy see earlier examples t = g

Time of flight for object dropped from height h 3. vx = 0.800 m/s Height of table = 0.960 m vx = 0.800 m/s dx = vx t t = ? 0.950 m Use dy = viy t + ½ g t2 dx = ? Time of flight for object dropped from height h 2 dy 2 h t = t = g g

3. vx = 0.800 m/s h = 0.960 m dx = vx t t = ? 2 h t = g 2 ( 0.950 m ) = t = 0.440 s = 9.8 m/s2 Then dx = vx t = ( 0.800 m/s )( 0.440 s ) dx = 0.352 m

4. dx = 52 m h = 122 m (a) Time of flight = ? 2 h t = g 2 ( 122 m ) = 9.8 m/s2 t = 5.0 s

4. dx = 52 m h = 122 m (b) vx = ? dx = vx t t t dx 122 m vx = = t t = 5.0 s dx = vx t 52 m t t dx 122 m vx = = t 5.0 s vx = 24 m/s

5. t = 4.5 s dx = 36 m (a) vx = ? dx = vx t t t dx 36 m vx = = t 4.5 s vx = 8.0 m/s

5. t = 4.5 s dx = 36 m (b) dy = ? Vertically, vi = 0 vx = 8.0 m/s (b) dy = ? dy = ? t = 4.5 s Vertically, vi = 0 dy = vi t + ½ g t2 36 m = 0 + ½ g t2 = ½ g t2 = ½ ( - 9.8 m/s2 )( 4.5 s )2 dy = - 99 m So cliff is 99 m high

6. A projectile is fired with a velocity of 196 m/s at an angle of 60.0o with the horizontal. (a) Find the vertical and horizontal velocities Draw horizontal and vertical components 196 m/s vy Use SOH-CAH-TOA 60o vx

SOH-CAH-TOA vy vx vx : vy : adj opp cos 60 = sin 60 = hyp hyp vx 196 196 m/s vy 60o vx vx : vy : adj opp cos 60 = sin 60 = hyp hyp vx 196 vy 196 (196) 0.500 = (196) 0.866 = 196 196 vx = 98 m/s vy = 170 m/s

(b) Find the time the projectile is in the air Projectile takes parabolic path to the ground 170 m/s As in earlier case, gravity is unaffected by horizontal motion 98 m/s Time of flight depends only on initial vertical velocity

(b) Find the time the projectile is in the air Projectile takes parabolic path to the ground 170 m/s As in earlier case, gravity is unaffected by horizontal motion 98 m/s Time of flight depends only on initial vertical velocity Time of flight will be the same for this projectile and for a projectile shot straight up with an initial velocity of 170 m/s

As in earlier cases, consider only 1st half of flight : vf = 0 vi = 170 m/s g = - 9.8 m/s2 vf = 0 t = ? As in earlier cases, consider only 1st half of flight 170 m/s : vf = 0 vf = vi + g t 98 m/s 0 = vi + g t - vi - vi - 170 m/s t = = 17.3 s - vi = g t - 9.8 m/s2 g g Multiply by 2 to get total flight time: - vi t = tt = 2 ( 17.3 s ) = tt = 34.6 s g

(c) Find the horizontal distance the projectile travels (this is called the range) Again, horizontal motion is affected only by horizontal velocity t = 34.6 s 170 m/s 98 m/s dx = ? dx = vx t = ( 98 m/s )( 34.6 s ) dx = 3390 m

(d) Find the maximum height the projectile reaches Height reached depends only on initial vertical velocity and time to reach the highest point 170 m/s vi = 170 m/s g = - 9.8 m/s2 t = 17.3 s (time of 1st half) 98 m/s dy = vi t + ½ g t2 = ( 170 m/s )( 17.3 s ) + ½ ( - 9.8 m/s2 )( 17.3 s )2 = 2940 m + ( - 1470 m ) dy = 1470 m

7. A projectile is fired at an angle of 53.0o with the horizontal at a velocity of 200 m/s. (a) Find the time the projectile is in the air 200 m/s vy Use SOH-CAH-TOA to find components of velocities 53o vx vx : vy : adj opp cos 53 = sin 53 = hyp hyp vx vy 200 200 (200) 0.602 = (200) 0.799 = 200 200 vx = 120 m/s vy = 160 m/s

Use vertical velocity to find time of flight vf = 0 vi = 160 m/s g = - 9.8 m/s2 t = ? vf = 0 ( for 1st half ) 160 m/s vf = vi + g t 120 m/s 0 = vi + g t - vi - vi - 160 m/s t = = 16.3 s - vi = g t - 9.8 m/s2 g g Multiply by 2 to get total flight time: - vi t = tt = 2 ( 16.3 s ) = tt = 32.6 s g

(b) Find the horizontal distance the projectile travels (the range) t = 32.6 s 160 m/s 120 m/s dx = ? dx = vx t = ( 120 m/s )( 32.6 s ) dx = 1960 m

8. A golf ball is hit at an angle of 45o with the horizontal at a velocity of 52 m/s. Find the range. 52 m/s 45o dx = ? dx = vx t Need to find vx and the time of flight First, find the velocity components using SOH-CAH-TOA

vy vx vx : vy : adj opp cos 45 = sin 45 = hyp hyp vx vy 52 52 (52) 52 m/s vy 45o vx vx : vy : adj opp cos 45 = sin 45 = hyp hyp vx vy 52 52 (52) 0.707 = (52) 0.707 = 52 52 vx = 36.8 m/s vy = 36.8 m/s

Use vertical velocity to find time of flight vf = 0 vi = 36.8 m/s g = - 9.8 m/s2 t = ? vf = 0 ( for 1st half ) 36.8 m/s vf = vi + g t 36.8 m/s 0 = vi + g t - vi - vi - 36.8 m/s t = = 3.75 s - vi = g t - 9.8 m/s2 g g Multiply by 2 to get total flight time: - vi t = tt = 2 ( 3.75 s ) = tt = 7.50 s g

dx = vx t = ( 36.8 m/s )( 7.50 s ) dx = 276 m dx = ? t = 7.50 s