Question

Problem (8):

Answer 1

Answer:

x=6 m, y= 3 m

Explanation:

Projectile Motion

It's the type of motion that experiences an object launched near the Earth's surface and moves along a curved path under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g the acceleration of gravity, then the components of the velocity at a given time t are:

$v_x=v_o\cos\theta$

$v_y=v_o\sin\theta-g.t$

And the components of the displacement are:

$x=v_o\cos\theta .t$

$\displaystyle y=v_o\sin\theta.t-\frac{g.t^2}{2}$

The projectile of the problem is fired at θ=53° with an initial speed of vo=10 m/s.

Thus, after 1 second, the components of the displacement are:

$x=10\cos 53^\circ \cdot 1$

$x\approx 6~m$

$\displaystyle y=10\sin 53^\circ\cdot 1-\frac{9.8\cdot 1^2}{2}$

$y\approx3~m$

Answer: x=6 m, y= 3 m

None of the options match this answer.