A ball weighing 1 kg was thrown from the Earth’s surface with an initial velocity of 20 m / s at an angle of 30 ° to the horizon. Consider that the air resistance when the ball moves is negligible. To what is the maximum height of the balloon? How has the potential energy of the ball changed during the ascent?
m = 1 kg.
g = 10 m / s ^ 2.
V0 = 20 m / s.
∠α = 30 °.
Only the gravity force m * g, directed vertically downward, will act on the thrown body.
Therefore, the body will horizontally move uniformly at a speed Vx, the value of which is found by the formula: Vx = V0 * cosα.
The body will move vertically with the acceleration of gravity g and the initial velocity Vу0 = V0 * sinα until it comes to a complete stop, and then it will move downward. The maximum height of the body hmax will be when the vertical component of the velocity is Vу = 0.
hmax = (V0 * sinα) ^ 2/2 * g.
hmax = (20 m / s * sinα30 °) ^ 2/2 * 10 m / s ^ 2 = 5 m.
Answer: the maximum height of the ball will be hmax = 5 m.
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