Question

How must a ball be projected from a height of 4 ft so as just to clear a wall 13 ft high, distance 15 ft
in a horizontal direction and a ditch 5 ft wide other side of the wall? (g=32 ft/s)
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Answer 1

Answer:

58/√3 with an angle of 46.4°

Explanation:

How must a ball be projected from a height of 4 ft so as just to clear a wall 13 ft high, distance 15 ft in a horizontal direction and a ditch 5 ft wide other side of the wall

When Vertical distance = 13 ft - 4 ft = 9 ft

at that time Horizontal Distance should be

15 ft  & 15 + 5 = 20 ft

Let say Its thrown with Velocity V at angle α

VCosα T₁ = 15

VCosα T₂ = 20

Vertical Distance  = VSinαT₁ - (1/2)gT₁² =  VSinαT₂ - (1/2)gT₂²  = 9 ft

=> 15Tanα - 16* 15²/V²Cos²α  = 20Tanα - 16* 20²/V²Cos²α = 9

Tanα =  (9  +  16* 15²/V²Cos²α)/15  = (9 + 16* 20²/V²Cos²α)/20

=> 36 + 14400/V²Cos²α  = 27 + 19200V²Cos²α

=> 9 = 4800/V²Cos²α

=> V²Cos²α = 4800/9

=>  V²Cos²α = 1600/3

=> Vcosα = 40/√3

using this

Tanα = 1.05

=> α = 46.4°

& Cosα = 1/1.45

V = 58/√3

58/√3 ft/s with an angle of 46.4°