Kinematics - 7 - 2D General Launch Angle

3.9 Objects Launched at an Angle: • Forms a parabola • Has both initial vertical & horizontal velocity • To calculate the components of velocity:  v1  v1x v1 y  v1x = v1 cos  v1 y = v1 sin Strategy: • Where possible, start by finding vx, v1y • Remember organization; horizontal and vertical variables kept separate (table!) Side note! • IF(!) object lands at same level as launch point, can find time to the apex (v2y = 0) and multiply by two for the total time. • Only works if object lands at same level (if dy=0)! Ex #1 20m/s 25o A soccer player kicks the ball with a velocity of 20.0m/s at 25.0o above the horizontal. a) How long does it take for the ball to reach its max height? b) What distance does the ball travel horizontally? c) What is the ball’s velocity at the moment it touches the ground?   v1 y = v1 sin v1 x = v1 cos vx = 20 cos25 v1 y = 20 sin 25 v x = 18.1262m / s v1 y = 8.45237m / s o o a) ∆d x X v2 y = v1 y + a y t Y =? Vx = 18.1262 m/s V1y = 8.45237 m/s 0 = 8.45237 + (−9.8)t V2y = 0 m/s ay = -9.80 m/s2 ∆t = ? b) ∆t = ? t = 0.862487s t = 0.862s Double the time back to ground! X Y ∆dx = ? ∆dy = 0 m Vx = 18.1262 m/s V1y = 8.45237 m/s ay = -9.80 m/s2 ∆t = 1.72497 s ∆t = 1.72497 s d x = vx t d x = (18.1262)(1.72497) d x = 31.3m v2 y = v1 y + a y t c) X Y ∆dx = 31.3 m ∆dy = 0 m Vx = 18.1262 m/s V1y = 8.45237 m/s V2y = ? ay = -9.8 m/s2 v2 y = 8.45237 + ( −9.8)(1.72497 v2 y = −8.45237m / s ∆t = 1.72497 s ∆t = 1.72497 s v x = 18.1262m / s Same two components as initial velocity, except downward facing  v = 20.0m / s [25.0o below the horizontal] (Pythagorean theorem and tangent confirms this) Ex #2 9m/s 35o From a roof 8.00m high, a ball is thrown with a velocity of 9.00m/s at 35.0o above the horizontal. a) How long does it take for the ball to reach the ground? b) What is the ball’s velocity at the moment it touches the ground? c) What is the max height reached by the ball?   v1 y = v1 sin v1 x = v1 cos vx = 9 cos35 o v x = 7.3724m / s v1 y = 9 sin 35 o v1 y = 5.1622m / s X a) ∆d Y =? ∆dy = -8.00 m Vx = 7.3724 m/s V1y = 5.1622 m/s x 1 d y = v1 y t + a y t 2 2 1 2 − 8 = 5.1622t + ( −9.8)t 2 V2y = ? t = 1.9088s t = 1.91s ay = -9.80 m/s2 b) ∆t = ? ∆t = ? (v ) = (v ) + 2a d (v ) = (5.1622) + 2(−9.8)(−8) 2 7.3724 m/s θ v2 2y 2 -13.5443 m/s 2 1y y y 2 2y v2 y = −13.5443m / s v2 = 15.4 m/s [61.4° below horizontal] (v ) = (v ) c) 2 2y X 2 1y + 2a y d y Y ∆dx = ? ∆dy = ? Vx = 7.3724 m/s V1y = 5.1622 m/s V2y = 0 m/s ay = -9.8 m/s2 (0) 2 = (5.1622) + 2( −9.8)(d y ) 2 d y = 1.360m Ex #3 22.0m/s 35o A ball is thrown with a velocity of 22.0m/s at 35.0o above the horizontal and it lands on a shelf which is 6.00m higher than the point where the ball was released. a) What is the max height reached by the ball? b) What is the horizontal displacement of the ball when it lands on the shelf?   v1 y = v1 sin v1 x = v1 cos v x = 22 cos 35 o v x = 18.0213m / s v1 y = 22 sin 35 o v1 y = 12.6187m / s a) ∆d X x Vx = 18.0213 m/s Y ∆dy = ? V1y = 12.6187 m/s V2y = 0 m/s ay = -9.80 m/s2 ∆t = ? (v ) = (v ) 2 2y ∆t = ? (0) 2 2 1y + 2a y d y = (12.6187) 2 + 2( −9.8)(d y ) d y = 8.12m b) X Y ∆dx = ? ∆dy = 6.00m Vx = 18.0213 m/s V1y = 12.6187 m/s 1 d y = v1 y t + a y t 2 2 V2y ay = -9.80 m/s2 ∆t = ? ∆t = ? t = 0.6292s t = 1.9460s d x = vx t 1 2 6 = 12.6187t + ( −9.8)t 2 4.9t − 12.6187t + 6 = 0 2 d x = (18.0213)(1.9460) d x = 35.1m 1 Summary of time solutions of d y = v1 y t + a y t 2 2 Vertical disp. If dy = 0 Time Which to use? t1 = 0 , t2 > 0 If dy < 0 t1 < 0 , t 2 > 0 If dy > 0 t1 > 0, t2 >0, t2 > t1 t2 is time at end of motion (t2/2 is time to max height) t2 is time at end of motion t2 is time at end of motion If solving total dx, use t2 in all cases HW: pg 257 #1-3 3.10 Projectile Range: • Another word for horizontal distance (“range” is dx) 2 ( v1 ) sin 2 range = − ay • Varies as a function of the launch angle – Zero at 0° or 90°, Maximum range 45° • Equal range for angles equidistant to 45° – (ex. 30° & 60° have same range) • ONLY VALID FOR SAME-LEVEL PROBS! Ex #4 Find the maximum range of a cannonball (launched from ground level, landing on ground) fired at a speed of 200.0 m/s. 2 v1 = 200.0m / s ( v1 ) sin 2 Range = − 2 ay a y = −9.80m / s d x = ? ( 200.0) =− 2 dx sin 2(45) − 9.80 Ex #1 (again) 20m/s 25o A soccer player kicks the ball with a velocity of 20.0m/s at 25.0° above the horizontal. - What distance does the ball travel horizontally? v1 = 20.0m / s 2 ( v1 ) sin 2 Range = −  = 25.0 a y = −9.80m / s d x = ? ay 2 dx 2 ( 20.0) sin(50.0) =− − 9.80 d x = 31.3m WILL NOT WORK FOR EXAMPLE 2!!! HW: pg 257 #4-6