A projectile is fired with an initial speed of 35.0 m/s at an angle of 40.0 abov

Question

A projectile is fired with an initial speed of 35.0 m/s at an angle of 40.0 above the horizontal. The object hits the ground 10.0 s later.

Part A

How much higher or lower is the launch point relative to the point where the projectile hits the ground?

Express a launch point that is lower than the point where the projectile hits the ground as a negative number.

Part B To what maximum height above the launch point does the projectile rise? Part C What is the magnitude of the projectile's velocity at the instant it hits the ground? Part D What is the direction (below +x) of the projectile's velocity at the instant it hits the ground?

Explanation / Answer

A)

along vertical

dy = voy*t + 0.5*ay*t^2

yf-yi = (35*sin40*10) - (0.5*9.81*10^2)

0 - yi = -265.52 m

yi = 265.52

launch point is higher at 265.52 m

+++++++++++

B)

at maximum height vy = 0

vy^2 - voy^2 = 2*ay*dy

0 - (35*sin40)^2 = -2*9.81*dy

dy = 25.6 m

C)

vy = voy + ay*T

vy = (35*sin40) - (9.81*10) = -75.6 m/s

vx = vox = vo*cos40 = 35*cos40 = 26.8 m/s

v= sqrt(75.6^2+26.8^2) = 80.21 m/s <<---------answer

++++++++

D)

direction = tan^-1(Vy/x) = 70.5

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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