Range, maximum height, and speed.

Question

A projectile is fired from ground level with an initial speed of 500 m/sec and an angle of elevation of 30 degrees. Use that the acceleration due to gravity is 9.8 m/sec^2.

(a) Find the range in meters.

(b) Find max height in meters.

(c) Find speed when the projectile hits the ground in m/sec.

Answer 1

y(t)=gt²/2+v_y0t + y₀ y₀ = 0

v_y0 = 500sin(30) = 250

v_x0= 500cos(30) = 250√3

x(t) = v_x0t - air_resistance ==> No air resistance given

Find t when y = 0 or ground level

-9.8t²/2 + 250t = 0

t = 0 and 51 seconds

Range = x(51) = 433(51) = 22083 m

Maximum height is when y'(t) = 0

y'(t) = gt+v_y0 =-9.8t+250 t = 250/9.8 = 25.5 seconds

y(25.5) = 3188.8 m

Max speed:

y'(51) = -9.8(51)/2+250 = -249.8 Speed = |y'(51)| = 249.8 m/s

x'(51) = 250√3= 433 m/s