Question

A ball is thrown towards a wall 10 m away with a velocity of 40m/s. At what two angles must it be thrown if it is to enter a hole on the wall at a height of 7m from the ground, neglecting wind effects​

Answer 1

Answer: 88.2 degrees, 40.1 degrees

Explanation:

We need to calculate the time using the provided values ( Range which is the horizontal component and the maximum height given which is the vertical component ). And from both, we derived an equation that could help us resolve the angle for the motion.

Answer 2

The horizontal velocity component is constant, so we can use the equation of motion:

d = v₀t + 0.5at²

where d is the distance to the wall (10 m), v₀ is the initial velocity (40 m/s), t is the time of flight, and a is acceleration due to gravity (-9.8 m/s²).

Since the ball is thrown at a height of 0 m, the vertical velocity component can be calculated using the equation of motion:

h = v₁t + 0.5at²

where h is the height of the wall (7 m) and v₁ is the initial vertical velocity.

Solving the two equations simultaneously for t:

10 = 40t + 0.5(-9.8)t²

and

7 = v₁t + 0.5(-9.8)t²

Squaring the second equation:

49 = (v₁t)² + 0.5(-9.8)t⁴

Substituting v₁t = 49 - 0.5(-9.8)t⁴ into the first equation:

10 = 40t + 0.5(-9.8)t²

Solving for t:

t = 1.33 s

Now, we can use t to find the vertical velocity component:

v₁ = -14.85 m/s

Finally, we can find the two angles of launch using the equation:

θ = sin⁻¹ (v₁ / v₀)

θ = sin⁻¹ (-14.85 / 40) = 68.59° and 111.41°

These two angles correspond to the two possible solutions for the problem, neglecting wind effects.

• In order to find the two angles at which the ball must be thrown to enter a hole on the wall at a height of 7m, we need to consider the ball's initial velocity and the time it takes to travel the 10m distance to the wall.
• The horizontal velocity component is constant, so we can use the equation of motion:
• d = v₀t + 0.5at²
• where d is the distance to the wall, v₀ is the initial velocity, t is the time of flight, and a is acceleration due to gravity (-9.8 m/s²).
• Since the ball is thrown at a height of 0 m, the vertical velocity component can be calculated using the equation of motion:
• h = v₁t + 0.5at²
• where h is the height of the wall and v₁ is the initial vertical velocity.
• Solving the two equations simultaneously for t, we can determine the time of flight, then use that time to calculate the vertical velocity component. The angle of launch can then be found using the equation:
• θ = sin⁻¹ (v₁ / v₀)
• This gives us two angles for which the ball will hit the wall at a height of 7m. Note that this solution neglects wind effects and other factors that may affect the ball's trajectory.

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