Two objects are thrown simultaneously from the ground, object 1 straight up and

Question

Two objects are thrown simultaneously from the ground, object 1 straight up and object 2 at an angle of $ heta$ = 32 degrees to the horizontal. The initial velocity of each object is 21.4 m/s. Choose g = 9.8 m/s^2 to answer the following questions. Neglect air resistance and assume both objects started from the same horizontal level (ground).

A)How high (vertically) is object 1 from the ground after t = 0.38 s?

B)How high (vertically) is object 2 from the ground after t = 0.38 s?

C)How far (horizontally) is object 2 from the initial point after t = 0.38 s?

D)What distance (linear) separates the two objects at t = 0.38 s?

E)What is the time of flight (ie, total time elapsed when it hits the ground) of object 1?

F)What is the time of flight of object 2?

Explanation / Answer

A) initial velocity of object 1 = 21.4 m/s upward

g = 9.8 m/s2

t = 0.38 s

S = u t - 0.5*gt2 = 21.4*0.38 - 0.5*9.8*0.382 = 7.42444 m

So height at time t = 0.38 s for object 1 is 7.42444 m

B) Vertical velocity initially for object 2 = 21.4 sin 32 = 11.34 m/s

S = u t - 0.5*gt2 = 11.34*0.38 - 0.5*9.8*0.382 = 3.60164 m

So height at time t = 0.38 s for object 2 is 3.60164 m

C) Horizontal velocity of object 2 = 21.4 * cos 32 = 18.148 m/s

time = 0.38 s

Horizontal distance covered by object 2 in 0.38 s = 18.148 * 0.38 = 6.896 m

D) Horizontal seperation at 0.38 s = 6.896 m

Vertical seperation at 0.38 s = 7.42444 - 3.60164 = 3.8228 m

So linear seperation = sqrt ( 6.896^2 + 3.8228^2) = 7.885 m

E) time of flight of object 1 = 2*vertical velocity initial/g = 2*21.4/9.8 = 4.367 s

F) time of flight of object 2 = 2*vertical velocity initial/g = 2*11.34/9.8 = 2.314 s

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