Physics, 24.05.2021 01:00 barbiegirllover

A projectile is launched from the ground with an initial velocity of 12m/s at an angle of 30° above the horizontal. The projectile lands on a hill 7.5m away. The height at which the projectile lands is most nearly

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Answer from: biaxialpower789

$1.78\:\mathrm{m}$

Explanation:

Kinematics equations used:

$v_f=v_i+at,\\\\\Delta y = v_it+\frac{1}{2}at^2,\\\\v_f^2=v_i^2+2a\Delta x$

Other equations used:

$\Delta x =v_i\cdot t,\\\text{-basic trigonometry for right triangles},\\\text{-application of physics concepts}$

By finding the horizontal component of the projectile's initial launch, we can find how long it will take for the projectile to reach the hill.

Since the projectile is launched at 12 m/s 30 degrees above the horizontal, we can use basic trig for a right triangle to find the horizontal component of the launch:

$\cos 30^{\circ}=\frac{x}{12},\\x=12\cos 30^{\circ}\approx 10.3923048454\:\text{m/s}$

Now we can use $\Delta x =v_x\cdot t$ to find the time it will take to reach the hill:

$7.5=10.3923048454\cdot t,\\t=\frac{7.5}{10.3923048454}\approx 0.72168783648\:\text{s}$

To find the total flight time the projectile would take if it were to land back to the same height it started out with, we can use the following kinematics equation:

v_f=v_i+at

We'll use basic trig for a right triangle again to find the vertical component of the velocity at launch:

$\sin 30^{\circ}=\frac{y}{12},\\y=12\sin 30^{\circ}=6\:\text{m/s}$ . (recall $\sin 30^{\circ}=\frac{1}{2}$ )

Now plugging this value in:

*Note: the projectile will hit the ground with a vertical velocity equal in magnitude but opposite in direction than the projectile's initial vertical velocity at launch

$-6=6+(-9.8\cdot t),\\t=\frac{-12}{-9.8},\\t=1.22448979592\:\text{s}$

^This is the total flight time the projectile would take if it were to land back to the same height it started with. Since acceleration is constant, we can divide this time by two to find the time it would take to reach maximum height.

$1.22448979592\div 2=0.61224489795\:\text{s}$

Thus, it will take $0.61224489795\:\text{s}$ to reach maximum height, meaning that it will take $0.72168783648 - 0.61224489795=0.10944293853\:\text{s}$ to land on the hill after reaching maximum height. Now that we've found this, we can use the following kinematics equation to find the vertical displacement from the projectile's max height to the hill:

$\Delta y =v_it+\frac{1}{2}at^2$

At maximum height, the projectile's vertical velocity will be zero. Therefore, v_i=0 and we have:

$\Delta y =\frac{1}{2}at^2$

Plugging in values , we have:

$\Delta y =\frac{1}{2}\cdot 9.8\cdot 0.10944293853^2=0.05869100829\:\text{m}$

This means that the hill is $0.05869100829\:\text{m}$ lower than the max. height the object will reach.

To find the height of the hill from the ground, we will use the following kinematics equation:

$v_f^2=v_i^2+2a\Delta x,\\-6^2=0+2\cdot-9.8\cdot\Delta x,\\36=-19.6\Delta x,\\\Delta x =\frac{36}{-19.6},\\\Delta x =-1.83673469388\:\text{m}$

Therefore, the max. height that the object will reach is $1.83673469388\:\text{meters}$ .

Thus, the hill's height is $1.83673469388-0.05869100829=1.77804368559\:\text{m}\approx\boxed{1.78\:\mathrm{m}}$ .

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