A mortar* crew is positioned near the top of a steep hill. Enemy forces are char

Question

A mortar* crew is positioned near the top of a steep hill. Enemy forces are charging up the hill and it is necessary for the crew to spring into action. Angling the mortar at an angle of Theta= 50.0 degress, the crew fires the shell at a muzzle velocity of 226 feet per second. How far down the hill does the shell strike if the hill subtends an angle Phi=34 degrees, from the horizontal?

How long will the mortar shell remain in the air?

How fast will the shell be traveling when it hits the ground?

I've convert 226 ft to 68.9 m/s. And I've found my X components are 68.9cos(50) = 44.3

Y components are 68.9sin(50) 52.8

Please help! Thank you!

Explanation / Answer

y = h + x

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.