A projectile is shot at an angle of #pi/4 # and a velocity of # 7 m/s#. How far away will the projectile land?

Answer 1

#x_"max"=4,99 " m"#

#v_i:"initial velocity"# #t_"elapsed":"elapsed time for projectile"# #v_x:"x component for initial velocity"# #x_"max"=v_x*t_"elapsed"# #v_x=v_i*cos alpha# #t_"elapsed"=(2*v_i*sin alpha)/g#

#x_"max"=v_i*cos alpha*(2*v_i*sin alpha)/g# #2*sin alpha*cos alpha=sin 2 alpha# #alpha=pi/4" "2alpha=pi/2" "sin(pi/2)=1# #x_"max"=(v_i^2*sin2 alpha)/g#

#x_"max"=(7^2 *1)/g# #x_"max"=49/(9,81)# #x_"max"=4,99 " m"#

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Answer 2

Nora Arzate

To find the horizontal range of the projectile, you can use the formula:

Range = (initial velocity)^2 * sin(2 * angle) / gravity

Given: Initial velocity (v) = 7 m/s Angle (θ) = π/4 radians Acceleration due to gravity (g) = 9.8 m/s²

Range = (7)^2 * sin(2 * π/4) / 9.8 Range = 49 * sin(π/2) / 9.8 Range = 49 * 1 / 9.8 Range ≈ 5 meters

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.