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jek_recluse [69]
3 years ago
12

How does the range of the pumpkin change if its initial velocity is tripled (keeping the angle fixed and less than 90∘)? How doe

s the range of the pumpkin change if its initial velocity is tripled (keeping the angle fixed and less than )? The pumpkin's range is nine times as far. The pumpkin's range is eighteen times as far. The pumpkin's range is three times as far.
Physics
1 answer:
Deffense [45]3 years ago
5 0

Answer:

The pumpkin's range is nine times as far.

Explanation:

Given,

In the first case,

Let a pumpkin is thrown with initial velocity u with an angel theta above the horizontal axis.

Therefore the range of the pumpkin is,

R\ =\ \dfrac{u^2sin2\theta}{g}\,\,\,\,\,\,\,\,\,\,\,\,eqn(1)

Now in the second case

Initial velocity of the pumpkin is three times the first case,

\therefore U\ =\ 3u

Let R' be the new range of the pumpkin.

New range of the pumpkin is,

\therefore R'\ =\ \dfrac{U^2sin2\theta}{g}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn(2)

From eqn (1) and (2), we get,

\therefore R'\ =\ \dfrac{(3u)^2sin2\theta}{g}\\\Rightarrow R'\ =\ \dfrac{9u^2sin2\theta}{g}\\\Rightarrow R'\ =\ 9 R

Hence the pumpkin's range is nine times of the initial case.

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3 years ago
A ball is thrown straight up from the ground with an unknown velocity. It reaches its highest point after 5.5 s. With what veloc
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3 years ago
A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0
1 year ago
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