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Akimi4 [234]
3 years ago
13

As a science fair project, you want to launch an 800 g model rocket st raight up and hit a horizontally moving target as it pass

es 30 m above the launch point. The rocket engine provides a constant thrust of 15.0 N. The target is approaching at a speed of 15 m/s. At what horizontal dista nce between the target a nd the rocket should you launch
Physics
1 answer:
fredd [130]3 years ago
4 0

Sum the forces in the y (upward) direction

\sum F_y = ma

F_t +F_w = ma

15N - (0.800kg)(9.81m/s^2) = 0.800kg * a

a = 8.94 m/s^2

Applying the kinematic equations of linear motion we have that the displacement as a function of the initial speed, acceleration and time is

s = v_0t + \frac{1}{2} at^2

30 = 0 +\frac{1}{2} (8.94) t^2

t = 2.59 s

Again through the kinematic equation of linear motion that describes velocity as the change of displacement in a given time, we have to

v = \frac{d}{t} \rightarrow d = vt

d = (15m/s)(2.59s)

d = 38.85m

Therefore the horizontal distance between the target and the rocket should be 38.83m

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A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet
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1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

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Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

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