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alisha [4.7K]
2 years ago
8

classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 128 feet per

second from an initial height of 112 feet off the ground, then the height of the projectile, h , in feet, t seconds after it's shot is given by the equation: h= -16t^2+128t+112 Find the two points in time when the object is 147 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places).
Physics
1 answer:
ddd [48]2 years ago
6 0

Answer:

t_1 = 0.28 s

t_2 = 7.72 s

Explanation:

Given that height of the projectile as a function of time is

h = -16 t^2 + 128 t + 112

here we know that

h = 147 ft

so from above equation

147 = -16 t^2 + 128 t + 112

16 t^2 - 128 t + 35 = 0

now by solving above quadratic equation we know that

t_1 = 0.28 s

t_2 = 7.72 s

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F=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(6 \cdot 10^{-6}C)(2 \cdot 10^{-6}C)}{(0.1 m)^2}=10.8 N


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Particle q2 wants to move in the direction of the force acting on it. The direction of the force depends on the relative sign of the two charges: like charges attract each other, opposite charges repel each other. In this case, the two charges are both positive, so they repel each other and q2 tends to move away from particle q1.

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