You throw a stone upward from the top of a building at an angle of 25â° to the horizontal and with an initial speed of 15 m/s. if the stone is in flight for 3.0 s, how tall is the building? how far from the base of the building does the stone.

11.025 m

87.75 m

Explanation:

Time of flight(T) of a projectile = 2U(sin∅)/g

Where U = initial Velocity, g = acceleration due to gravity, ∅= angle of projection.

Make ∅ the subject of the the equation,

∅ = sin⁻¹[(T× g)/2U]

Where U = 15m/s, T= 3.0 s, g = 9.8 m/s²

∅ = sin⁻¹[(3 × 9.8)/(2×15)]

∅ = sin⁻¹(29.4/30)

∅= sin⁻¹(0.98) = 78.52°

Using the formula for maximum height of a projectile

H = U²sin²∅/2g

H = 15²(sin²78.52)/2 × 9.8

H = 225(0.98 × 0.98)/19.6

H = 216.09/19.6

H = 11.025 m

Range (R) = U²sin2∅/g

R = 15²sin(2×79.52)/9.8

R = 225(0.39)

R =87.75 m

∵ the building is = 11.025 m tall and the base of the building is 87.75m away from where the stone landed.

i think they are all correct

explanation:

they seem good

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.