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A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40° above the horizontal. How far above or below its original level will the ball strike the opposite wall?

Respuesta :

Blacklash

Answer:

Ball hit the tall building 50 m away below 10.20 m its original level

Explanation:

Horizontal speed = 20 cos40 = 15.32 m/s

Horizontal displacement = 50 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    50 = 15.32 t + 0.5 x 0 x t²

     t = 3.26 s

Now we need to find how much vertical distance ball travels in 3.26 s.

Initial vertical speed  = 20 sin40 = 12.86 m/s

Time = 3.26 s

Vertical acceleration = -9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²

    s = -10.20 m

So ball hit the tall building 50 m away below 10.20 m its original level

onyebuchinnaji

The distance the ball will strike the opposite wall is 32.79 m.

Time of motion of the ball

The time of motion of the ball from the given height is calculated as follows;

h = vsinθ(t) + ¹/₂gt²

50 = 20 x sin(40)t + 0.5(9.8)t²

50 = 12.86t + 4.9t²

4.9t² + 12.86t - 50 = 0

solve the quadratic equation using formula method,

t = 2.14 s

Horizontal distance of the ball

The horizontal distance of the ball from the initial position is calculated as follows;

X = vcosθ(t)

X = 20 x cos(40) x 2.14

X = 32.79 m

Thus, the distance the ball will strike the opposite wall is 32.79 m.

Learn more about horizontal distance here: https://brainly.com/question/24784992

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