Question

A projectile is thrown from the top of a platform of height h = 10 m, with velocity v with magnitude 14 m/s and forming an angle 35 deg from the horizontal direction. The projectile is at a horizontal distance d1 away from the launch point when it is back at the same height h, and the horizontal distance d2 from the launch point when it hits the ground.
What is the distance d1?

Answer 1

The horizontal distance traveled by the projectile is 9.51 m.

The given parameters;

• height of the platform, h = 10 m

• velocity of the projectile. v = 14 m/s

• angle of projection, = 35 degrees

The  horizontal distance of the projectile from the point of projection is range of the projectile.

The horizontal component of the projectile is calculated as;

$v_0_x = v_0 \times cos(\theta)\\\\v_0_x = 14 \ m/s \times cos(35)\\\\v_0_x = 11.47 \ m/s$

The time of flight of the projectile,

$h = v_0_yt + \frac{1}{2} gt^2\\\\10 = (14 \times sin(35)) t + 0.5 \times 9.8t^2\\\\10 = 8.03 t + 4.9t^2\\\\4.9t^2 + 8.03t - 10 = 0\\\\solve \ the quadratic \ equation \ using \ formula \ method;\\\\a = 4.9, \ b = 8.03, \ c = -10, \\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-8.03 \ \ +/- \ \ \sqrt{(8.03)^2 - 4(4.9\times -10)} }{2(4.9)}\\\\t = 0.83 \ s$

The horizontal distance of the projectile from the point of projection is calculated as;

$d_1 = v_0_x \times t\\\\d_1 = 11.47 \ m/s \times 0.83\\\\\d_1 = 9.52 \ m$

Thus, the horizontal distance of the projectile from the point of projection (d₁)  is 9.52 m.

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