Could anyone answer these question step by step. 2. A cannonball is launched wit

Question

Could anyone answer these question step by step.

2. A cannonball is launched with an initial velocity of 10.0 m/s at an angle of 19.0° above the horizontal from a cliff that is 75.00 m tall. Neglect friction inside the cannon and air resistance. Find the following: a. The time that it will take for the cannonball to land. 3. 5 sec b. The magnitude of the velocity at landing 35 472 dwdX 38.5 V notv c. The distance from the base of the cliff to the point that the cannonball lands. 36.8 m un Nox Sano Vof t 3.54

Explanation / Answer

Q1.
u = 10 m/s
angle = 19 deg
h = 75 m

a)
t = 2v0sin(theta)/g

=> t = (2*10*sin19)/9.8
=> t = 0.67 s

b)
v = sqrt(u^2 + 2h(-g))
=> v = sqrt(100 + 2(-75)(-9.8))
=> v = 40 m/s

c)
R = u^2*sin2(theta) / g
=> R = (100*2sin19cos19)/9.8
=> R = 6.3 m

Pls post other ques separately

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---