Question

Problem (8):

A projectile is fired at an angle of 53° above the horizontal with initial speed of 10.0 m/s. After
one second, what is X and Y components of the position of that projectile?
A60m 20 m
B. 6.0 m. 8.0 m
C. 3.0 m, 6.0 m
D.4.0 m, 3.0 m​

Answer 1

x=6 m, y= 3 m

Step-by-step 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-(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-(9.8\cdot 1^2)/(2)`

`y\approx3~m`

Answer: x=6 m, y= 3 m

None of the options match this answer.