You throw a ball from ground level with a mass of 1 kilogram upward with a velocity of v = 24 m/s on Mars, where the force of gravity is g = -3.711 m/s².
A. Approximate how long the ball will be in the air on Mars.
B. Approximate how high the ball will go.
Answer (2)
To solve this problem, we need to determine two things:
A. Approximate how long the ball will be in the air on Mars.
B. Approximate how high the ball will go on Mars.
We can use the equations of motion for constant acceleration to solve these. Since the ball is moving under the influence of gravity alone after it's thrown, we can apply these equations.
A. Time the ball will be in the air:
We use the formula for the time of flight of a projectile when it returns to the same vertical level:
t = ∣ g ∣ 2 v
Where:
v = 24 m/s (initial velocity)
g = − 3.711 m/s 2 (acceleration due to gravity on Mars)
Substitute in the values:
t = 3.711 2 × 24 ≈ 12.93 seconds
So, the ball will be in the air for approximately 12.93 seconds.
B. Maximum height the ball will reach:
To find the maximum height, we use the formula for vertical displacement when the final velocity at the top is zero:
h = 2∣ g ∣ v 2
Substitute the known values:
h = 2 × 3.711 2 4 2
h = 7.422 576 ≈ 77.60 meters
Thus, the ball will reach a maximum height of approximately 77.60 meters.
By breaking down the problem into these steps, we've calculated both the total time the ball is in the air and the maximum height it reaches on Mars.
The ball will be in the air for approximately 12.93 seconds and reach a maximum height of about 77.60 meters on Mars.
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