Question

A pistol that fires a signat flare gives it an initial velocity (mUZZIe velocity) of 153 m/s at an angle of 55.2 degree above the horizontal. you can ignore air resistance. Find the flare's maximum height if it is fired on the level salt flats of Utah. Find the distance from its firing point to its landing point if it is fired on the level salt flats of Utah. Find the flare's maximum height if it is fired over the flat Sea of Tranquility on the Moon, where 9 = 1.67m/s2.Find the distance from its firing point to its landing point if it is fired over the flat Sea of Tranquility on the Moon, where 9 = 1.67m/s2.

Answer 1

Hi,

To solve this problem you should remember some kinematic formulas. As the question is telling us to ignore the air resistence the only force over the signal flare is the weight, therefore we only have acceleration (g in this case) in the y axis.

Before answering we should know the velocity in each axis:

v_0x = cos(&)*l v₀ l = cos(55.2)*153 = 87.3 m/s

v_0y = sen(&)*l v₀ l = sen(55.2)*153 = 125.6 m/s

Besides, they are telling us that we are working in Utah (in certain parts), so we are going to assume that g = 9.8 m/s² (standard value on Earth's surface)

Part A

The maximum height can be found when you assume that at that point the velocity in the y axis is cero.

v_y² = v_0y² - 2gy; so, y_max = v_0y² / (2g) = (125.6)² / (2*9.8) = 804.9 m

Part B

Assuming that the landing ponit of the signal flare is reached by it after going up to y_max and going down the same distance we have the following.

v_y = v_y0 - gt; if v_y = 0;   t_max = v_y0/g = (125.6/9.8) = 12.8 s; then the maximum distance in the x axis (distance to the landing point) is:

x_max = v_x0*(2t_max) = 87.3*2*12.8 = 2234.9 m

The next parts are done in the same way, the only difference is that we must use a new value of g.

Part C

y_max = v_0y² / (2g) = (125.6)² / (2*1.67) = 2828.2 m

Part D

t_max = v_y0/g = (125.6/1.67) = 75.2 s

x_max = v_x0*(2t_max) = 87.3*2*75.2 = 13129.9 m

I hope it helps.