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3 years ago

The acceleration vector of a particle in projectile motion ________.

Physics

1 answer:

Alex73 [517]3 years ago

4 0

Answer:

Points downward, and its magnitude is 9.8 m/s^2

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.

- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.

The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.

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Read 2 more answers

A mountain climber stands at the top of a 50.0 m cliff hanging over a calm pool of water. The climber throws two stones vertical

Sonbull [250]

Answer:

a. $t=2.997\ s$ is the time after which the two stone hit the water surface.

b. $u_2=1.998\ m.s^{-1}$

c. $v_1=31.3688\ m.s^{-1}$ & $v_2=31.3686\ m.s^{-1}$

Explanation:

Given:

height of the cliff, $h=50\ m$

time gap between the projection of stones, $\Delta t=1\ s$

initial velocity of first stone, $u_1=+2\ m.s^{-1}$

Here positive sign means that the stone is thrown vertically downward in the direction of gravity.

Using equation of motion:

$h=u_1.t+\frac{1}{2} g.t^2$

$50=2t+4.9t^2$

$t=2.997\ s$ is the time after which the two stone hit the water surface.

using the equation of motion for second stone:

$h=u_2.t+\frac{1}{2} g.t^2$

$50=u_2\times 2.997+4.9\times 2.997^2$

$u_2=1.998\ m.s^{-1}$

Final velocity of the stone at the instant of touching the water surface:

$v_1^2=u_1^2+2.g.h$

$v_1^2=2^2+2\times 9.8\times50$

$v_1=31.3688\ m.s^{-1}$

$v_2^2=u_2^2+2g.h$

$v_2^2=1.998^2+2\times 9.8\times 50$

$v_2=31.3686\ m.s^{-1}$

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