Please everything step by step. And do what it asked.

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 2s (from the time of release) determine the instantaneous velocity of the bomb. Determine the instantaneous velocity after 3s and using a vector diagram show the acceleration acting on the bomb.

2. A hot air balloon is rising vertically at a speed of 48m/s. At a 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 velocity when it reaches the ground?

3. At one instant a pool ball is traveling 1/2 m/s[N30E], 3s later it is going 0.3m/s [N45W]. What was the average acceleration in this interval of time?

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