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

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3 years ago

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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

Physics

2 answers:

mihalych1998 [28] · Answer 1 · 2022-11-04T16:17:25+03:00

mihalych1998 [28]3 years ago

5 0

here is your answer hope it's helpful for you thankyou I

faust18 [17] · Answer 2 · 2022-11-04T17:56:56+03:00

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:

$s = v_{x}t$