Please help me please. Instructions and formulas are giving down below.

Projectile Summary Sheet: Projectile Case Sketch of case Time of Flight Components Range Maximum height Instantaneous Impact velocity velocity (vector (vector addition) addition) Case 1 Vis tf = v2d/g Launch speed Vi R = Vit Hmax = d Vinst = Vv + Vh added Same as Vinst, -Launched where d = Vv = 0 Where d is the as vectors Vimpact = Vv + Vh with t = t horizontally vertical height Vh = Vi launch height Where Vv = gt -Lands at a between Note: Vh is And Vh = launch speed Where Vv = gtf different level launch and constant and Vv is And vh = launch speed landing always changing - R Same as Vinst Case 2 tt = 2visine Launch speed Vi R = vi sin20 hmax = Vizsinze Vinst = Vv + Vn -Launched at an and launch angle 28 Where Vinst = Vv + Vh Vv = Visine + gt Where t=tf angle E or -Lands at same Vv = Visine R = Vitf And Vn = Vicose Vv = Visine + gtf level Vh = Vicose And Vn = Vicose R Note: Vh is constant and Vv is always changing Same as Vinst Case 3 tup =0 -Visine Launch speed Vi R = Vit: d = vit + 1/2at Vinst = Vv + Vi -Launched at an and launch angle with v1 = Visine Where Vinst = Vv + Vh Vv = Visin0 + gt Where t=t angle E t = tup and a = g -Lands at a tdown = V2d/g gives max height And Vn = Vicose Vv = Visin0 + gtf Vin Vv = Visine different level Le where d = Vh = Vicose above launch And Vn = Vicose **Choose a launch height Note: Vhis level. direction to be plus distance it constant and Vv is O positive* * flies upward. always changing d = 1/2at2 with t = tdown and azg RAlternate Method to Solve Case 2 Projectile Problems Note: The equations derived below are only useful in solving Case 2 Projectile Problems, that is projectiles launched from and landing at the same elevation. Since the projectile is being launched at an angle 9, we know that the launch speed vi can be broken into a horizontal component and a vertical component. Vh = Vlcose and sz V.sine where V: is the initial launch speed and e is the launch angle measured to horizontal. Time of Flight From what you read, you realize that the time of flight of a projectile depends on the vertical component of the launch speed and gravity. Also from grade 11, you know that the speed an object is launched vertically will be the same speed it is going on the way down. V2V1 a Recall: t = if we designate up as positive and down as negative, V1 = Visine (vertical component of VI) V2 = - Vlslne (Same speed, opposite direction) a = -9.8 m/s2 or -g V2V1 t : a t _ ~Visin3 "Visine '9 2VisinB t = '9 2Visin6 t = .9 So for a CASE 2 projectile this formula will allow you to solve for the time of flight by plugging in the launch speed for V. , the angle for e, and 9.8 for g presuming the projectile is on earth, Pro lecTile MoTion Problems and VecTor AcceleraTion 1, A bomb is dropped from an airplane flying horizonTally aT a speed of 600km/h aT a heighT of 490m. When and where does The bomb sTrike The ground? AfTer a Time of 23 (from The Time of release) deTermine The insTanTaneous velociTy of The bomb. DeTermine The insTanTaneous VelociTy offer 33 and using a vecTor diagram show The acceleraTion acTing on The bomb. 2, A hoT air balloon is rising verTically (IT a speed of 48m/s. AT (1 heighT of 640m, a ball is projecTed horizonTally wiTh a speed of 30m/s. When and where will The ball reach The ground? WhaT is The resulTanT velbciTy when iT reaches The ground? 3. AT one insTanT a pool ball is Traveling -~ m/s[N30E], 35 laTer H is going 0.3m/s [N45W]. WhaT was The average acceleraTion in This inTerval of Time? 4. A bomb is dropped by a 81 bomber flying level 11' 400m/s [N]. The bomb sTrikes The ground 5 seconds laTer. a) How high was The plane? b) WhaT was The insTanTaneous velociTy JusT before impacT? c) How far NorTh of The release poinT does The bomb sTrike? d) WhaT was The overall displacemenT of The bomb? 5. In a waTerballoon fighT, one Team uses a slingshoT ThaT can launch The balloons aT 50 m/s. a) AT whaT angle(s) should They launch The balloon To hiT a TargeT 100m away? b) If They launched a balloon aT 20 To The horizon, when and where will iT hiT? WhaT will iTs velociTy aT impacT be? 6. A golf ball is hiT wiTh an iniTial speed of 100m/s aT an angle of 40. DeTermine: ' a) Time of flighT b) Range c) Maximum heighT d) WhaT is The insTanTaneous veloci'ly aT impacT? \fWhen the the bomb will hit the ground, $10/8 x 22 10. 6 m O the speed of the plane, Goormin . 6000-1500 m's 3600 D. 8 8 - 9.025 Range = Vill - 83 3 4.4 m .. it will strive of 9:028 from 5 3 3 41.4 me dill . . instonton eaus velocity often is = Vit gt - = 16 6 4 + ( 24 9. 8 )- 1656.6 m/g adten 38 similarly = 1606 4 mls ~ BEG DOC 11.13 5 D. 8 ange = ( 30 * 11.43 )m = 342.86 m . it will reach the ground at 11.AB sec and 342. 86 m dam aug= 115. 06 7MIS - 0-1062 0.206 2 - 0.37x Cmlgb The height was d 2 - 1 x 3. 8 x 28 = 1228-8 m instantaneous velocity a = (9. 8 + 8 ) + 400 FEARS mls for = (5 x 100 ) m = 2000 m D Z ( 2050)+ ( 122.872 = 2004 m 202 sin 20 P = zy 20 = Sin 1 60 * 9. 8 -1 (301 2 . - 23.080 0 = 11 . 50