Question

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 passes 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

Answer 1

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

Answer 2

The horizontal distance between the target and the rocket should be 38.83m.

Kinematic equations of linear motion:

Since There is launch an 800 g model rocket straight up and hit a horizontally moving target as it passes 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.

Now the sum of forces in the y or upward direction is

f_t + f_w = ma

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

a = 8.94m/s^2

Now here we have to determine the time

We know that

s = v_ot + 1/2at^2

30 = 0 + 1/2(8.94)t^2

t = 2.95s

Now the distance is

d = vt

= 15 * 2.59

= 38.85m

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