of 22 m/s in a direction making an angle of 25 degrees upward with the horizontal. a. What is the magnitude (angle) of the velocity of the ball when it hits the ground? b. How long was the ball in air? c. What is the horizontal range of the object?

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A: Given Data, Initial speed of the shuttlecock(v)= 18 m/s Angle above the horizontal (θ)=380

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A: Approach to solving the question:Projectile formulaDetailed explanation:

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A: (a) velocity of the object vt=∫adt=∫0,-gdt=0,-gt+u0,v0=0,-9.8 m/s2t+30,6 m/s=30,-9.8t+6 m/s

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A: Since you have posted multiple questions, we will provide the solution only to the first question…

Q: Two spheres are launched horizontally from a 1.2 mm -high table. Sphere AA is launched with an…

A: As spheres are launched horizonatally so time depends only on height. t=2hg