An object is launched from a platform (3300m above groundlevel ) at an angle of

Question

An object is launched from a platform (3300m above groundlevel) at an angle of 35 degrees, this object lands 9400away from the platform at ground level. At whatinitial speed was this object launched at?

My main problem was the platform, I do not understand how to goabout any question that has an elevated launch or lands above theinitial launching level.

Any help would be appreciated! Thanks.

Explanation / Answer

the trick is to realize that the object velocity has a horizontalcomponent Vx and a vertical component Vy at launch, and it takesthe same time t for the horizontal component Vx to travel 9400m asthe vertical component Vy to start from an elevation of 3300m, goup until it reaches the highest point and then come back down allthe way to the ground level. So we start with Vx and Vy: Vx = V cos350 Vy = V sin350 Remember V, the intial launch speed, is unknown. In time t (the total flight time), horizontal motion: Vx t = V cos350 t = 9400 >>>equation A vertical motion: Vy t - (1/2) g t2 = -3300 (this is freefall equation; y-axis is upwards; displacement is negative) The 2nd equation can be rewritten as - (1/2) g t2 + V sin350 t + 3300 = 0>>> equation B Now the steps are: 1. From equation A, solve for Vt 2. Subtitute that in equation B and solve for t. 3. Once you know t, use equation A to find V. Hope this helps!

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