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Elza [17]
3 years ago
12

Which statement about projectile motion is true?

Physics
2 answers:
Shalnov [3]3 years ago
4 0

Answer:

D. The hang time of a projectile is greatest for a vertical launch.

Explanation:

As we know that range of projectile is given by the formula

R = \frac{v^2sin(2\theta)}{g}

so here as we increase the angle till 45 degree then the range will decrease but if we increase the angle more than 45 degree then the range will start decreasing

Also we know that time of flight is given by the formula

T = \frac{2vsin\theta}{g}

since it depends on the vertical component of the motion so we can say that hang time can be determine by vertical components of the motion

So maximum hang time is possible only when angle of projection is 90 degree

so correct answer will be

D. The hang time of a projectile is greatest for a vertical launch.

Svetllana [295]3 years ago
3 0

C. The range of a projectile increases with an increase in the angle of launch.

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Answer:

1) d

2) 5 m/s

3) 100

Explanation:

The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:

(i) x=\frac{1}{2}at^2+v_0t+x_0

The equation for velocity v and a constant acceleration a is:

(ii) v=at+v_0

1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) a = \frac{v -v_0}{t}

(iv) x = \frac{v-v_0}{2t}t^2+v_0t + x_0

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

(v) x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t

2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}

3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}

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Answer:

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Explanation:

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