Question

A projectile is shot from the edge of a cliff 2.57 x 10^-16Em above the ground level with an initial speed of 4.32 Km/min at an angle of 66.0^0 with the horozontal. (a) Determine the time taken by the projectile to hit the ground. (b) The range of the projectile from the base of the cliff. At a certain point K 6.5 x 10^-2 Km from the ground, find (c) the vertical and horizontal components of its velocity, (d) the magnitude of the velocity, (e) the angle made by the velocity vector with the horizontal, (f) the maximum height above the cliff top reached by the projectile. Select one: a. a: 16.6 s, b: 477 m, c: Vkx=29.3 m/s; Vky= 88.9 m/s, d: Vk= 95.0 m/s, e: 72.0^0, f: 230 m b. a: 16.6 s, b: 486 m, c: Vkx=29.3 m/s; Vky= 90.0 m/s, d: Vk= 94.6 m/s, e: 72.0^0, f: 221 m c. a: 15.7 s, b: 487 m, c: Vkx= 30.1m/s; Vky= 99.9 m/s, d: Vk= 92.6 m/s, e: 72.0^0, f: 202 m d. a: 14.3 s, b: 478 m, c: Vkx=27.7 m/s; Vky= 87.0 m/s, d: Vk= 94.6 m/s, e: 71.0^0, f: 263 m

          A projectile is shot from the edge of a cliff 2.57 x 10^-16Em above the ground level with an initial speed of 4.32 Km/min at an angle of 66.0^0 with the horozontal. (a) Determine the time taken by the projectile to hit the ground. (b) The range of the projectile from the base of the cliff. At a certain point K 6.5 x 10^-2 Km from the ground, find (c) the vertical and horizontal components of its velocity, (d) the magnitude of the velocity, (e) the angle made by the velocity vector with the horizontal, (f) the maximum height above the cliff top reached by the projectile.

Select one:
a. a: 16.6 s, b: 477 m, c: Vkx=29.3 m/s; Vky= 88.9 m/s, d: Vk= 95.0 m/s, e: 72.0^0, f: 230 m
b. a: 16.6 s, b: 486 m, c: Vkx=29.3 m/s; Vky= 90.0 m/s, d: Vk= 94.6 m/s, e: 72.0^0, f: 221 m
c. a: 15.7 s, b: 487 m, c: Vkx= 30.1m/s; Vky= 99.9 m/s, d: Vk= 92.6 m/s, e: 72.0^0, f: 202 m
d. a: 14.3 s, b: 478 m, c: Vkx=27.7 m/s; Vky= 87.0 m/s, d: Vk= 94.6 m/s, e: 71.0^0, f: 263 m

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A projectile is shot from the edge of a cliff 2.57 x 10^-16Em above the ground level with an initial speed of 4.32 Km/min at an angle of 66.0^0 with the horozontal. (a) Determine the time taken by the projectile to hit the ground. (b) The range of the projectile from the base of the cliff. At a certain point K 6.5 x 10^-2 Km from the ground, find (c) the vertical and horizontal components of its velocity, (d) the magnitude of the velocity, (e) the angle made by the velocity vector with the horizontal, (f) the maximum height above the cliff top reached by the projectile.

Select one:
a. a: 16.6 s, b: 477 m, c: Vkx=29.3 m/s; Vky= 88.9 m/s, d: Vk= 95.0 m/s, e: 72.0^0, f: 230 m
b. a: 16.6 s, b: 486 m, c: Vkx=29.3 m/s; Vky= 90.0 m/s, d: Vk= 94.6 m/s, e: 72.0^0, f: 221 m
c. a: 15.7 s, b: 487 m, c: Vkx= 30.1m/s; Vky= 99.9 m/s, d: Vk= 92.6 m/s, e: 72.0^0, f: 202 m
d. a: 14.3 s, b: 478 m, c: Vkx=27.7 m/s; Vky= 87.0 m/s, d: Vk= 94.6 m/s, e: 71.0^0, f: 263 m

Added by Toni C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Hubert Agamasu

11/15/2021

Step 1

4.32 km/min = 4.32 * 1000 m/60 s = 72 m/s Then, we can find the initial horizontal and vertical components of the velocity: V_0x = V_0 * cos(θ) = 72 m/s * cos(66°) = 29.3 m/s V_0y = V_0 * sin(θ) = 72 m/s * sin(66°) = 64.3 m/s (a) The time taken by the Show more…

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A projectile is shot from the edge of a cliff 2.57 x 10^-16Em above the ground level with an initial speed of 4.32 Km/min at an angle of 66.0^0 with the horozontal. (a) Determine the time taken by the projectile to hit the ground. (b) The range of the projectile from the base of the cliff. At a certain point K 6.5 x 10^-2 Km from the ground, find (c) the vertical and horizontal components of its velocity, (d) the magnitude of the velocity, (e) the angle made by the velocity vector with the horizontal, (f) the maximum height above the cliff top reached by the projectile. Select one: a. a: 16.6 s, b: 477 m, c: Vkx=29.3 m/s; Vky= 88.9 m/s, d: Vk= 95.0 m/s, e: 72.0^0, f: 230 m b. a: 16.6 s, b: 486 m, c: Vkx=29.3 m/s; Vky= 90.0 m/s, d: Vk= 94.6 m/s, e: 72.0^0, f: 221 m c. a: 15.7 s, b: 487 m, c: Vkx= 30.1m/s; Vky= 99.9 m/s, d: Vk= 92.6 m/s, e: 72.0^0, f: 202 m d. a: 14.3 s, b: 478 m, c: Vkx=27.7 m/s; Vky= 87.0 m/s, d: Vk= 94.6 m/s, e: 71.0^0, f: 263 m

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Transcript

00:03 Hello, so we have the height of the cliff being 2 .57 times 10 to the power minus 16 of an eta meter.

00:26 Because it is 10, 10 to the power 18.

00:36 So this gives us 257 meters as the height of the cliff.

00:47 The initial velocity is 4 .3 to.

00:55 Kilometers over 60 seconds.

01:01 So that is 72 meters per second.

01:06 So our projectile is fired from the top of a cliff.

01:19 That is on 257 meters high.

01:26 We found initial velocity of 72 meters per second making an angle of 7 meters per second making an angle of 7.

01:36 6 to 6 degrees to the horizontal.

01:40 So first of all we're going to find the time it takes to hit the ground.

01:45 We use the formula s equals s note plus vyt and it's half t squared projectile formula.

02:01 S is the final displacement when it hits the ground.

02:04 Since the ground is a reference point when it's on the ground displacement is zero.

02:09 S not is any original displacement above the ground before the motion started.

02:14 So in this case, s not is the displacement 257 or the height of the projector at time t equals you.

02:25 Vy is the vertical component of velocity which is which is vertical component of velocity is 72, 6 to 6.

02:40 Horizontal component is 72 called 66 and so this 72 sign 6.

02:51 Minus 5 ,9 .8 t squared.

02:56 This comes down to a quadratic equation in t, for which we have two answers.

03:10 T equals 16 .6 seconds and also minus 3 .2 seconds.

03:18 Of course, we take the positive one and is the negative.

03:25 For b, we are to find a range.

03:27 The range is simply the horizontal velocity...

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