A ball is thrown at an original speed of 8.0 m/s at an angle of 35° above the horizontal.If there is no air,what is the speed of the ball when it returns to the same horizontal level?
A)4.0 m/s
B)8.0 m/s
C)16 m/s
D)9.8 m/s

Question

Quizplus · Accepted Answer

The initial vertical velocity can be calculated using the given angle and initial speed: VY = V0 sin(θ) = 8.0 sin(35°) = 4.58 m/s
The horizontal velocity remains constant throughout the motion, so the final horizontal velocity is the same as the initial horizontal velocity: VX = V0 cos(θ) = 8.0 cos(35°) = 6.55 m/s
When the ball reaches the same horizontal level, its vertical displacement is zero. We can use the equation of motion for vertical motion to find the final speed:
Vf^2 = VY^2 + 2aΔy
where a is the acceleration due to gravity (-9.8 m/s^2) and Δy is the vertical displacement (which is zero in this case)
Vf^2 = 4.58^2 + 2(-9.8)(0) = 21.0 m^2/s^2
Vf = √21.0 = 4.58 m/s (rounded to two significant figures)
Therefore, the answer is B) 8.0 m/s.

A Ball Is Thrown at an Original Speed of 8