Question

PLEASE HELP !! ILL GIVE BRAINLIEST A projectile is launched from the top of a 15 m tall building at a speed of 25ms at an angle of 50° above the horizontal. The horizontal distance from the base of the building to where the projectile lands is most nearly

Answer 1

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Answer 2

This question involves the concepts of projectile motion and the equations of motion.

The horizontal distance from the base of the building to where the projectile lands is most nearly "73.6 m".

First, we will apply the first equation of motion to the upward motion of the projectile to find the time of upward motion.

$v_{fy}=v_{iy}+gt_{up}$

where,

$v_{fy}$ = final speed's vertical component at top = 0 m/s

$v_{iy}$ = initial speed's vertical component = (25 m/s)Sin 50° = 19.15 m/s

g = acceleration due to gravity = - 9.81 m/s² (upward motion)

$t_{up}$ = time taken for upward motion = ?

Therefore,

$0\ m/s = 19.15\ m/s + (-9.81\ m/s^2)(t_{up})\\\\t_{up} = \frac{19.15\ m/s}{9.81\ m/s^2}\\\\t_{up} = 1.95\ s$

Now, we will apply the second equation of motion to the upward motion of the projectile to find the height reached.

$h_{up} = v_{iy}t_{up}+\frac{1}{2}gt_{up}^2\\\\h_{up} = (19.15\ m/s)(1.95\ s)+\frac{1}{2}(-9.81\ m/s^2)(1.95\ s)^2\\\\h_{up} = 37.34\ m - 18.65\ m\\h_{up} = 18.69\ m$

He,ce the  total heigh (h) will be:

h = 18.69 m + 15 m = 33.69 m

Now, we will apply the second equation of motion to the downward motion of the projectile to find the time taken during downward motion.

$h = v_{iy}t_{down}+\frac{1}{2}gt_{down}^2$

where,

g = 9.81 m/s² (downward motion)

$v_{iy}$ = initial velocity's vertical component at top = 0 m/s

Therefore,

$33.69\ m = (0\ m/s)t_{down}+\frac{1}{2}(9.81\ m/^2)t_{down}^2\\\\t_{down} = \sqrt{\frac{(2)(33.69\ m)}{9.81\ m/s^2}}\\\\t_{down} = 2.62\ s$

Hence, the total time taken (t) by the projectile motion is:

$t = t_{up}+t_{down}\\t = 1.95\ s + 2.62\ s\\t = 4.57\ s$

Now, we will consider the horizontal motion of the projectile. Ignoring the air resistance, we can apply the uniform motion equation to the horizontal motion:

$s = v_{x}t$

where,

s = distance from the base of building to where projectile lands = ?

$v_x$ = horizontal component of launch velocity = (25 m/s)Cos 50° = 16.1 m/s

t = 4.57 s

Therefore,

s = (16.1 m/s)(4.57 s)

s = 73.6 m

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.