A person tosses a ball from the ground up into the air at an initial speed of 10 m/sec and an initial angle of 43◦ off the groun
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Answer:
F = -49.1 10³ N
Explanation:
Let's use the kinematics to find the acceleration the acceleration of the bullet that they tell us is constant
² = v₀² + 2 a x
Since the bullet is at rest, the final speed is zero
x = 11.00 cm (1 m / 100 cm) = 0.110 m
0 = v₀² + 2 a x
a = -v₀² / 2 x
a = -1320²/(2 0.110)
a = -7.92 10⁶ m / s²
With Newton's second law we find the force
F = m a
F = 6.20 10⁻³ (-7.92 10⁶)
F = -49.1 10³ N
The sign means that it is the force that the tree exerts to stop the bullet